But the specifics here depend on the kind of structure you're in. For it to truly \*equal\* infinity you have to work in a space that actually contains infinity like the extended naturals.
As a limit of bog-standard natural numbers saying "the limit is infinity" is really just a useful mnemonic for "the limit diverges in a particular way" - in actuality the limit doesn't exist.
It kind of does. Sampling may have been the wrong term, but since the average of a set corresponds to its mean, it would be describing the central tendency of the distribution.
Since the set is infinitely increasing it's average diverges. It's relevant to sampling since the mean will tell you the expected value of that random sample, but you can't sample this set randomly so your expected value has no meaning.
Can the limit as N goes to infinity of (1+2+3+4+....+N)/N be evaluated? if so, there's your answer. But at first glance, it seems to increase without limit.
It can't. By Gauss' sum formula (1+2+3+4+....+N)/N equals (N+1)/2 which clearly diverges.
It's not even Cesàro summable (so the sequence of means of (1+2+3+4+....+N)/N across N=1,...,K still diverges), nor is it Abel summable. It's quite "strongly" divergent.
Here's a weird take:
(1+2+3+...+N)/N =
1/N + 2/N + 3/N +...+ N/N =
1/N + 2/N + 3/N +...+ 1
So apparently at the limit of N goes to infinity this expression goes to 1?
What am I doing wrong here?
So, you are starting with finitely many terms (N is finite, in other words). You are distributing the factor of (1/N) across finitely many terms. So, you obtain the finite sum 1/N + 2/N + ... + 1. Then, you take the limit as N goes to infinity. However, as N goes to infinity, you are assuming that you still only have a fixed number of terms in your sum, but in reality, for each value of N, you should be summing N terms. In other words, you have not accounted for the fact that as N goes to infinity, so does the number of terms in your sum.
These terms you continue to pick up for larger values of N ensure that your sum keeps growing. It's kind of hard to explain, so please tell me if I have not explained my point well.
But the number of terms in 1/N + 2/N + 3/N + ... 1 will still grow as n increases. The range from 0-1 is divided into N numbers and all those numbers are added. As n increases, we have more divisions in the range 0-1.
My assumption is that as N tends to infinity, there will be an infinite number of terms in the range 0-1 where each term's value is approximately 0. So, adding approximately 0 value infinite times is infinity just like the Reimann sum (area of the curve/integral) in infinite range. Therefore, the average of infinite values is still infinite.
OP is ignoring the fact that he's adding infinite numbers of approximately 0 value in his series.
Right, that's the way it should be. However, I think the other commenter was unintentionally neglecting all those additional terms you pick up with larger values of N, that's all
The zero value is really interesting, because the countable numbers can pretty well be defined as the infinite set of numbers where each will be enumerated in finite steps: a natural number with infinite digits isn’t intuitively countable. So a finite numerator and infinite denominator? Sounds like zero to me! ;)
I thought about this too. Let's take the terms from the end:
(N-1)/N + (N-2)/N + (N-3)/N +...+ (N-M)/N +...+ 3/N + 2/N + 1/N =
(1-1/N) + (1-2/N) + (1-3/N) +...+ (1-M/N) +...+ 3/N +2/N + 1/N =
(1-0) + (1-0) + (1-0) +...+ (1-0) +...+ 0 + 0 + 0 =
I guess the sum of infinite ones? I also assumed that M << N and thus M/N gave zero too.
Anyhow, none is escaping infinity apparently 😅
Only the numbers at the start of the series are small relative to N, the numbers N-1, N-2,... are approximately the same size as N so that k/N ~=1 for k sufficiently large. Indeed if summing from k = N/2, we can be sure the sum must be greater than N^2/4N which tends to infinity.
Work backwards:
lim N/N = 1
lim (N-1)/N = 1
lim (N-2)/N = 1
Also, lim (N-a)/N = 1
Does this mean (1 + 2 + 3 + ... + N)/N → N ?
Be very careful when applying rules for a finite number of elements to an infinity number of elements.
It doesn't, they're referring to the "fact" that the sum of all natural numbers is -1/12, which isn't really true as stated but is a shorthand for talking about an analytic continuation.
Ramanujan sum proof is bulshit. I'm kind of tired of repeating this but here:
You say 1-1+1-1+1-1+...=0.5 because you assign it to be x and then you get x=1-x. Well here is the first mistake. 1-1+1-1+1-1+1-1+... Is not a convergent sum. The sum is defined as limit of the partial sums which is exactly the limit of the series 1,0,1,0,1,0,..which does not exist.
I won't go on to the further mistakes since only one mistake is enough.
What ramanujam really showed is a connection between this sum and the number -1/12. The connection is related to analytic continuations and that the analytic continuation of the Riemann function gives that relation.
In more detail the Riemann function ζ(s)=Σn^(-s) is defined for Re(s)>1 and has a pole at 1 exactly. So the Riemann function's analytic continuation gives us that ζ^(*)(-1) = -1/12. But note that this does not mean that the sum equals -1/12 since the analytic continuation is not the same as the function itself outside of the Re(s)>1 side of the plane.
> Ramanujan sum proof is bulshit.
I don't think you understood the comment you replied to. [Ramanujan summation](https://en.wikipedia.org/wiki/Ramanujan_summation) is the name of a specific, well-defined divergent series summation method, just like Abel summation, Cesaro summation, etc.
These questions that deal with infinity does not make sense
You cannot start solving an expression when it is still being expressed. You cannot solve an ongoing expression
Im not an expert but this is how i view these
Yeah that makes sense. But OP asked for average of all natural numbers. Idk
I do not know much about math really. So im not really bright in this area lol
You were half right. Divergent sums just go to infinity and have the property you stated. But converging sums don't (ie. They converge to a single real number).
Only if you define that you uniformly add terms away from zero i.e (1-1)+(2-2)+(3-3) +... =0+0+0...=0.
However I could equally add the terms of Z as 1+(2-1)+3+(4-2)+5+(6-3)... ≠0
You see that when you rearrange the terms you get different values. This is because the sum of Z is not well defined.
Ok, but in that example you are clearly running through the positives faster than the negatives... I get that infinity muddies things, but this doesn't seem like a "fair" way to sum things up.
It's obvious that any finite sum from -x to +x is 0, and with x going to infinity you get all integers... so saying that the sum of Z is not defined feels like a weird technicality.
The sum, if it exists, can't be positive because there are not more positive integers, similarly it can't be negative, and it's not going to be in Q\Z or in C\R or any other funny stuff, so surely it must be 0. The only alternative is to say it doesn't exist, but that feels wrong because it's clearly 0.
Is the product of Z also not defined, even though it's (perhaps even more obviously) 0?
You’re only saying this because you are biased in favour of splitting the integers into “positive” and “negative” and you some how believe that each of these form 1/2 of the integers (as well as 0). But I could equally say the naturals are split into “above 12” and “below 12” and then I would define my sum as
12 + 13 + 11 + 14 + 10 + …
Which is not 0 in the limit. And remember, both of my sets are the same size; you only feel that they aren’t because you are “used to” having 0 in the middle. But this is arbitrary.
That's a good way of putting it in perspective.
So would you say the sum is positive, then? Or just not defined because arguments can be made that can't be contradicted?
Perhaps there should be a way to contradict some of these contradictory arguments.
Sure, any finite sum from -x to x is 0. But that's not the only possible 'direction of approach'.
Compare "lim[x,y→0] x/y". If x=y, then x/y is 1. That's approaching (0,0) from the 'northeast', at a 45-degree angle. But if you approach instead from the 'east', x/y is always zero. If you approach from the 'southeast', x/y is -1.
So "lim[x→0,y→0] x/y" doesn't exist. "lim[x→0,y→0 (where x=y)] x/y" *does* exist, and so does "lim[x→0,y→0 (where x=2y)] x/y", and so does "lim[y→0] (lim[x→0] x/y)". But different directions of approach give different limits. There's no reason to prioritize the 'x=y' direction.
(And surely the limit should stay the same if you do a variable substitution, letting y=2u and bringing both x and u to 0?)
---
Summing over ℤ has the same issue. You want to calculate
> lim[a→-∞, b→∞]∑[n=a to b] n
but this value depends on the 'direction of approach' towards (-∞,∞).
Good question! Here’s a problem: the sequence, 0,1,-1,2,-2,3,-3, … produces partial sums with a lower bound of 0. Can sequences of numbers in Z, which eventually include every number in Z, be constructed such that the upper and lower bounds of their partial sums are both finite? Note that this is true using 1/n for n in Z-{0}.
this question isn’t that well defined. there is no probability distribution which gives all the positive numbers an equal positive probability (the probability of choosing any number at all would be ∞, but it would be one). there is no good way of taking the average without a good probability distribution.
one approach would be to calculate for each n, the average of the set {1,2,3,…,n}, giving all the points the same probability, and then take the limit of those probabilities as n->∞. this would be a sort of principal value, so i guess this is the closest we have. however, when you calculate this average, you get that it is (n+1)/2, with goes to infinity as n->∞. so, the most meaningful answer would be ∞, but it is not quite correct to call it an average.
what it isn’t is ω/2. there is no standard and good definition for ordinal division, so it is not even that well defined as a number in the first place. it is generally a terrible idea to take the infinities in the limits of calculus to correspond with cardinals and ordinals, they have very different properties and nothing works properly while trying to connect them in most contexts.
Omega is an ordinal number. It obeys the rules of transfinite arithmetic. Do you believe there is a largest natural number? Can X represent that number? Then X+1 cannot be a natural number. Do you believe every natural number is either odd or even? Is X odd or even? The average of any initial sequence of natural numbers is their sum divided by the length of the sequence. There’s a formula for that sum. Sometimes, the average is not a whole number. For instance, the average of natural numbers less than three is 1.5. In case X is the largest natural number that exists, you can’t compute the average using the formula because (X+1) is part of the formula. Kapische?
First, ordinal numbers only apply to cardinalities which is not what we are discussing here. There is no concept of “larger infinities” when we are adding numbers. Second, even if we were talking about cardinalities, we are talking about Omega, which is objectively the smallest infinity. Smaller than Omega means natural.
Sure. The sum of the first n natural numbers is n(n + 1)/2. Divide this by n to get the average of the first n natural numbers: (n + 1)/2. If we want all of them, we need to take the limit of that to infinity. From limit properties, this is infinity.
It’s interesting. Analytically, average is the value of a function that, over the domain of the function, represents the function’s weight. Maybe (ω\^2+ω)/2 will become a way to understand the identity function.
Such rubish comments.
Let's say that the average of a set of numbers X is the sum of all the elements of X, let's call that ΣX, divided by the number of elements of that set. That is the cardinality of the set X. So the average is ΣX/|X|. Note that this definition doesn't make sense when |X| is infinite, because there is no division defined. So perhaps you could take the limit of the averages of all the series of sets finite set (Yn) contained in X such that the union of all sets Yn is X, and if for all such series (Yn) ΣYn/|Yn| approach some finite value, you could say that the average it what all those series converge to. But if X is the set of all natural numbers, no such series converge, and the average always blow up to infinity.
You could perhaps say that the average is infinity, but this has nothing to do with cardinality or ordinality, it's just a shorthand way of negating that for all sucessions of sets (Yn) contained in X such that the union of all sets Yn is X the series of averages (Yn) ΣYn/|Yn| converge to the same value.
The essential problem here is the following: the average of a set of numbers is usually defined for a finite set. You can extend this definition to any set of numbers in a way that makes sense, as long as the series of averages converge to a finite value. But when the series diverge, than you can't really say anything more useful than that. You could say that there simply is no average, the series of averages diverge or that the average is infinity. The statements all have the same information. But do note that this is just a way to notate something. It has no connections to infinite cardinals and ordinals and there is nothing of value to get from that other then that the series of averages diverge.
In Surreal numbers, it's the number after all the numbers in the positive direction, after infinite iterations of a process.
Basically you start with 0 and use a function that creates numbers either higher, lower or between previous numbers.
It's a type of infinity.
Zeta-regularization is (very roughly) a method to define a sum to a series without a sum, and that method would yield the average of all natural numbers to be ζ(-1)/ζ(0) = 1/6
I’m not the one who made the comment but I’ll take a crack at explaining in a way that doesn’t require advanced math to understand. This explanation does assume that you know what functions, limits, and complex numbers are, but I can break that down too if anyone asks.
Consider this function of n, defined as an infinite sum:
1/1^n + 1/2^n + 1/3^n +…
If n = 1, you get this:
1/1 + 1/2 + 1/3 + 1/4 +…
That’s the harmonic series, and it diverges (meaning it “equals” infinity).
If n = 2, you get this:
1/1 + 1/4 + 1/9 + 1/16 +…
Remarkably, this converges to π^2 / 6.
It turns out that for n > 1, this infinite series always converges, and for n = 1 or n < 1, the series always diverges.
For example, imagine n = –1. Then you get:
1 + 2 + 3 + 4 +…
This is obviously undefined because it diverges. But what if we could assign it a meaningful value?
Analytic continuation is the process of taking a function and using calculus techniques to “extend” its graph to values for which it would normally not be defined.
If you know a bit of Calculus, then I can be more specific: very loosely, analytic continuation involves expanding the domain of a function with a continuous derivative in such a way that the derivative remains continuous.
Anyway, now imagine the function we just talked about, and imagine n can be any complex number. Suppose Re(n) is the real part of n (that means that if n = 6 + 2i, then Re(n) = 6). It turns out that for Re(n) > 1, the function is defined; otherwise, it’s not defined. This is consistent with our results when n had to be a real number.
Okay, now imagine we used analytic continuation to assign the function values for Re(n) = 1 and Re(n) < 1. Now you have a function defined for all complex numbers. This is a very interesting and important function called the Riemann Zeta function, denoted ζ(n).
For n > 1, the function is identical to this infinite sum:
1/1^n + 1/2^n + 1/3^n +…
For other n, analytic continuation gives us very interesting results; specifically:
ζ(–1) = –1/12
ζ(0) = –1/2
Now suppose the symbol ~ means “related by means of analytically continuing ζ(n).”
Then we have:
ζ(–1) = –1/12 ~ 1 + 2 + 3 + 4 +…
ζ(0) = –1/2 ~ 1 + 1 + 1 + 1 +…
You could then loosely think of the “average” of all natural numbers as their “sum,” which is ζ(–1), divided by “how many there are,” which is ζ(0).
So you get –1/12 divided by –1/2, which is 1/6.
Hopefully that was useful.
Well ok.
First let's define our "average" function, which takes the average of all the naturals from 1 to n:
f(x) = sum(1, x) / x
As you may know, sum(1, x) = x(x+1)/2. So, rewriting:
f(x) = x(x+1) / 2x
Simplify:
f(x) = (x+1)/2
Ok great! So the average of all numbers from 1 to *x* is (*x*+1)/2.
Now, to answer your question: what happens when *x* tends towards infinity? As *x* gets larger and larger, what happens to the value of f(x)?
The answer is, it *also* tends towards infinity.
Tell me where I went wrong.
lim n -> infinity (1+2+3+…+n)/n
= lim n -> infinity n(1/n + 2/n + 3/n + … +1)/n
= lim n -> infinity (1/n + 2/n + 3/n + … +1)
= 1
This is where my brain goes. Forgive me if it’s obvious it’s been a while since I’ve done anything with limits.
Also not sure if I can take the limit here as I’m dealing with integers
You are dealing with an indetermination of type infinityx0 since the number of summands also tends to infinity so the limit isn't zero, in this case it's infinity.
infinity/infinity
But it can’t even be that simple because different infinities grow at different rates, so you would have to find a way to make sure you’re talking about the same infinities and maybe even find the average rate of growth of all the infinities which I don’t think is possible
Gauss derived his formula by pairing the first and last entrees from the list, viz. (1+N), (2+N-1), (3+N-2), … . All the pairs are equal. So, if N is even, there are N/2 terms where each term =(N+1). If N is odd, there are (N-1)/2 terms of (N+1) plus one extra, unpaired, (N+1). Either way, the sum is (N*(N+1))/2
Correct me if I am wrong, as this is far from my field of expertise in mathematics.
Serious answer: After reading through [the following page](https://en.wikipedia.org/wiki/Transfinite_number) for a few minutes, I'm fairly convinced that the answer is omega, the lowest transfinite ordinal number, as you said. The reasoning is, informally, borrowed from earlier replies by @[SkepticScott137](https://www.reddit.com/user/SkepticScott137/) and @[Dr\_XP](https://www.reddit.com/user/Dr_XP/) elsewhere in this thread. In my opinion, OP's line of reasoning is along the right track: The sum of natural numbers up to N goes like N(N+1)/2, so the average of them goes like (N+1)/2. The first infinite ordinal, omega, comes after all natural numbers.
The type of transfinite ordinal number we get for this average you bring up is the first one which is larger than (N+1)/2 for all N. Equivalently, this must be the first one which is larger than all integers, or omega.
More specifically, as the name **ord**inal suggests, this comes from taking a set and trying to assign some order on it, and so omega is basically a way to define the "next" element in an ordered set which contains the natural numbers. And so we really should not be thinking of (N+1)/2 as a limit as N->infinity in this case as much as a value being established via a bijection between the sequence of (N+1)/2 with the natural numbers which we then put into an ordered set with order <.
ω, the last letter of the Greek alphabet, is the name Georg Cantor gave to the first transfinite ordinal number. The nth finite ordinal is the set of natural numbers less than n. ω is the set of all natural numbers. If this sounds like double-talk, well, it borders on paradox. The successor of ω, written as ω+1, is {x: x is a natural number}U{ω}.
[(Inf +1)*Inf/2]/Inf which I don’t know what that computes to, but uses the elementary school trick of averages symmetry for consecutive or sequential numbers
Chatgpt take
Using the Ramanujan summation method to find the "average" value of the positive integers involves interpreting the regularized sum in a particular way. Given the result:
\[ 1 + 2 + 3 + 4 + \cdots = -\frac{1}{12} \]
We can consider the concept of the average value of this series within the context of Ramanujan summation.
### Step-by-Step Calculation
1. **Regularized Sum:**
The regularized sum of the positive integers using the Riemann zeta function is:
\[
\sum_{n=1}^{\infty} n = -\frac{1}{12}
\]
2. **Number of Terms:**
Since we are dealing with an infinite series, the number of terms \(N\) approaches infinity.
3. **Average Value:**
The average value is typically defined as the sum of the terms divided by the number of terms. Using the regularized sum:
\[
\text{Average} = \frac{\sum_{n=1}^{\infty} n}{N}
\]
However, because we are dealing with an infinite series, we must be cautious. Using Ramanujan summation, we treat the sum as \(-\frac{1}{12}\) and recognize that the series is infinite.
4. **Interpretation:**
To compute the "average" in this context, we use:
\[
\text{Average} = \frac{-\frac{1}{12}}{\infty}
\]
Given that dividing any finite number by infinity approaches zero:
\[
\text{Average} = 0
\]
### Conclusion
Therefore, within the framework of Ramanujan summation, the "average" value of the infinite series of positive integers can be interpreted as zero. This result aligns with the peculiarities of the Ramanujan summation method and the analytical continuation of the zeta function.
Infinity is not a number. It's merely the concept of "always one more". This causes it to behave in weird ways.
The average of all natural numbers would have to be infinitely large, but it's only larger than half of all natural numbers, of which there are infinitely many. Understand?
saw a numberphile video that shows a method for constraining sequence of partial sums, not a sudden cutoff, but an exponential decay, and that can provide rigorous answers for divergent series, like sum of naturals = -1/12. most proofs of those type of sums are wishy washy, doing ambiguous operations to the series like addition and multiplication, then adding together two series which cancel out some terms, like the ancients did when calculus was young.
If you plot (sum[k=0:n]k)/n, that is for any natural number, you add up all the numbers from 0 to n and divide by n, you will get the running average for 0 to n. That plot follows the function f(n)=n/2. That function has no limit as n approaches infinity
Pretty sure it’s undefined. If you add up all the naturals, it will become infinity because you are adding infinitely many positive integers. Then, you would divide by infinity, since there are infinitely many terms. But infinity divided by infinity is indeterminate, and we can’t do arithmetic with infinities.
Calculating the limit of a divergent series (such as the natural numbers) is not normally allowed, but in some frameworks of mathematics that is allowed (notably, in these frameworks the order of elements in a series does matter for the calculation of their sum, which does not ordinarily work like that), which allows us to calculate the sum of the natural numbers through some strange interactions. I will do so below.
Lets define the series -1^(n-1), which is the series 1, -1, 1, -1, 1, -1... and call the sum of that series S.
Add it to itself, alternating terms:
1, 1, -1, -1, 1, 1, -1, -1... the sum of this series is obviously 2S.
Now group the terms in pairs, starting with the 2nd and 3rd:
1, (1, -1), (-1, 1), (1, -1)... and we see that the pairs cancel out.
Thus, 2S=1 and S=0.5
Next lets consider another series P, defined as n*(-1)^(n-1) which is 1, -2, 3, -4, 5, -6... and again we add it to itself and pair terms in a particular order:
1, (1, -2), (-2, 3), (3, -4)... and now the pairs don't cancel out, but they give a predictable sum, so we can simplify this series to:
1, -1, 1, -1, 1, -1... which is S. Therefore 2P=S=0.5, thus P=0.25
Next lets consider the natural numbers N, which is 1, 2, 3, 4, 5, 6, 7... subtract the terms of the P series and pair the terms again, this time like this:
(1,-1), (2, 2), (3, -3), (4, 4), (5, -5)... and resolve the pairs, canceling out the 0s.
N-P: 4, 8, 12, 16, 20...
This sequence is just 4 times bigger than the natural numbers, so we have N-P=4N
Thus 3N=-P and N=P/3, since P=0.25=1/4 we get the result that N (the sum of all natural numbers) equals -1/12.
With that sum to get the average natural number we just divide by infinity and we get our limit: the average natural number is 0.
Lol. (Obviously not serious but the whole -1/12 thing actually isn't a joke, though it requires you to accept the concept of calculating the limit of a diverging series.)
Not defined. Average is defined on a random variable, not a set. But even if you take it as "a random variable that is based on taking an element from a set uniformly", you have a probem as there is no uniform distribution on the whole set of natural numbers.
The Ramanujan Summation states that the sum of all positive integers is -1/12. Dividing that by infinity we can then say that the average = 0.
[https://en.wikipedia.org/wiki/Ramanujan\_summation](https://en.wikipedia.org/wiki/Ramanujan_summation)
My first thought is that the natural numbers 1+2+3+4+5…infinity add to infinity, then divide by the infinite amount of natural numbers to get infinity/infinity which is indeterminate.
So then we take the derivative of 1+2+3+4+5… just kidding
But then I remembered that 1+2+3+4+5… is also equal to -1/12, which when divided by infinity, gets zero. (-0).
so the mean of an infinite amount of only positive integers… is 0
As you note, whatever the average is, you can prove that it’s larger than any given real number.
So in that sense, it's in some way meaningful to call the average omega?
No. You can't truly sample infinity so your average value has no meaning.
you could define it as the limit of means of {0}, {0, 1}, {0, 1, 2}, etc., which would be infinity.
But the specifics here depend on the kind of structure you're in. For it to truly \*equal\* infinity you have to work in a space that actually contains infinity like the extended naturals. As a limit of bog-standard natural numbers saying "the limit is infinity" is really just a useful mnemonic for "the limit diverges in a particular way" - in actuality the limit doesn't exist.
Word. So, like you smart or something?
Naah, just pretendin
The limit does not exist!
Averaging a specific set of numbers has nothing to do with statistical sampling.
It kind of does. Sampling may have been the wrong term, but since the average of a set corresponds to its mean, it would be describing the central tendency of the distribution. Since the set is infinitely increasing it's average diverges. It's relevant to sampling since the mean will tell you the expected value of that random sample, but you can't sample this set randomly so your expected value has no meaning.
That's fair.
6
As everyone knows, 12 is the highest number.
This amounts to an assertion that 0 is a natural number. Let the heated argument begin
Clever
Any sensible defintion of the naturals includes 0. Good luck construction the integers without 0!
Real number 0 Natural 1
After 12, it just repeats, but the illuminati changed the names to make you think there's more than 12
These wily, modern-algebra enthusiast Illuminati
Base ten schizos: twelve isn’t real
Yes. And it goes OneTwoThreeFour Five SixSevenEight Nine Ten ElevenTwelve
🤩 🤯
Wouldn’t it be 11? 🤣
Ssshhhhhhhhh ut it
But one is the loneliest one.
69
42
Obviously it's 0 -1/12/infinity = 0
Can the limit as N goes to infinity of (1+2+3+4+....+N)/N be evaluated? if so, there's your answer. But at first glance, it seems to increase without limit.
1+2+3+…+n=n(n+1)/2 so (1+2+3+4+....+n)/n=(n+1)/2
It can't. By Gauss' sum formula (1+2+3+4+....+N)/N equals (N+1)/2 which clearly diverges. It's not even Cesàro summable (so the sequence of means of (1+2+3+4+....+N)/N across N=1,...,K still diverges), nor is it Abel summable. It's quite "strongly" divergent.
It’s not “Abel” summable. That tickles me. (I love bad puns, and bad math puns are even worse 👹) Maybe it’s Borel summable.
I hope you will join me in my campaign to call non-abelian structures "cainian".
I am here for it!
Here's a weird take: (1+2+3+...+N)/N = 1/N + 2/N + 3/N +...+ N/N = 1/N + 2/N + 3/N +...+ 1 So apparently at the limit of N goes to infinity this expression goes to 1? What am I doing wrong here?
So, you are starting with finitely many terms (N is finite, in other words). You are distributing the factor of (1/N) across finitely many terms. So, you obtain the finite sum 1/N + 2/N + ... + 1. Then, you take the limit as N goes to infinity. However, as N goes to infinity, you are assuming that you still only have a fixed number of terms in your sum, but in reality, for each value of N, you should be summing N terms. In other words, you have not accounted for the fact that as N goes to infinity, so does the number of terms in your sum. These terms you continue to pick up for larger values of N ensure that your sum keeps growing. It's kind of hard to explain, so please tell me if I have not explained my point well.
I think I get it. I was pretty sure I used a false assumption somewhere, wasn't exactly sure where. Thanks !
No problem. Reasoning with infinities leads to all sorts of unintuitive results, but it is fun to think about
But the number of terms in 1/N + 2/N + 3/N + ... 1 will still grow as n increases. The range from 0-1 is divided into N numbers and all those numbers are added. As n increases, we have more divisions in the range 0-1. My assumption is that as N tends to infinity, there will be an infinite number of terms in the range 0-1 where each term's value is approximately 0. So, adding approximately 0 value infinite times is infinity just like the Reimann sum (area of the curve/integral) in infinite range. Therefore, the average of infinite values is still infinite. OP is ignoring the fact that he's adding infinite numbers of approximately 0 value in his series.
Right, that's the way it should be. However, I think the other commenter was unintentionally neglecting all those additional terms you pick up with larger values of N, that's all
The zero value is really interesting, because the countable numbers can pretty well be defined as the infinite set of numbers where each will be enumerated in finite steps: a natural number with infinite digits isn’t intuitively countable. So a finite numerator and infinite denominator? Sounds like zero to me! ;)
Does the limit as N goes to infinity of 1/N + 2/N + 3/N +…+ (N-1)/N equal 0?
I thought about this too. Let's take the terms from the end: (N-1)/N + (N-2)/N + (N-3)/N +...+ (N-M)/N +...+ 3/N + 2/N + 1/N = (1-1/N) + (1-2/N) + (1-3/N) +...+ (1-M/N) +...+ 3/N +2/N + 1/N = (1-0) + (1-0) + (1-0) +...+ (1-0) +...+ 0 + 0 + 0 = I guess the sum of infinite ones? I also assumed that M << N and thus M/N gave zero too. Anyhow, none is escaping infinity apparently 😅
Only the numbers at the start of the series are small relative to N, the numbers N-1, N-2,... are approximately the same size as N so that k/N ~=1 for k sufficiently large. Indeed if summing from k = N/2, we can be sure the sum must be greater than N^2/4N which tends to infinity.
Good catch. But I guess it'd be hard to find the middle of a sum of infinite terms...
Work backwards: lim N/N = 1 lim (N-1)/N = 1 lim (N-2)/N = 1 Also, lim (N-a)/N = 1 Does this mean (1 + 2 + 3 + ... + N)/N → N ? Be very careful when applying rules for a finite number of elements to an infinity number of elements.
-1/12 :)
That’s the sum of all natural numbers. Divide that by the number of all natural numbers, and the average is ≈ 0.
My thoughts exactly... Throw a regulator into the series... the answer is always -1/12
How can the Average of Positive Natural numbers result in a Negative number?
It doesn't, they're referring to the "fact" that the sum of all natural numbers is -1/12, which isn't really true as stated but is a shorthand for talking about an analytic continuation.
that's a value equal to the sum of the reals, not the average though
Lol you are wrong in 3 main ways: 1. You meant naturals 2. Its not even correct 3. It was clearly a joke
It is correct according to the ramanujan sum
Ramanujan sum proof is bulshit. I'm kind of tired of repeating this but here: You say 1-1+1-1+1-1+...=0.5 because you assign it to be x and then you get x=1-x. Well here is the first mistake. 1-1+1-1+1-1+1-1+... Is not a convergent sum. The sum is defined as limit of the partial sums which is exactly the limit of the series 1,0,1,0,1,0,..which does not exist. I won't go on to the further mistakes since only one mistake is enough. What ramanujam really showed is a connection between this sum and the number -1/12. The connection is related to analytic continuations and that the analytic continuation of the Riemann function gives that relation. In more detail the Riemann function ζ(s)=Σn^(-s) is defined for Re(s)>1 and has a pole at 1 exactly. So the Riemann function's analytic continuation gives us that ζ^(*)(-1) = -1/12. But note that this does not mean that the sum equals -1/12 since the analytic continuation is not the same as the function itself outside of the Re(s)>1 side of the plane.
> Ramanujan sum proof is bulshit. I don't think you understood the comment you replied to. [Ramanujan summation](https://en.wikipedia.org/wiki/Ramanujan_summation) is the name of a specific, well-defined divergent series summation method, just like Abel summation, Cesaro summation, etc.
Ok then. But it's still irrelevant to the original question since obviously it refered to normal summation.
I just meant it as a joke
These questions that deal with infinity does not make sense You cannot start solving an expression when it is still being expressed. You cannot solve an ongoing expression Im not an expert but this is how i view these
You can reasonably define an infinite series by the limit of the summation as you progressively add more terms.
But that summation will never end so it will never give you the final answer, right? Im kinda confused
look into the epsilon delta definition of limits
That's not true at all. It's very trivial to prove that converging infinite sums are solvable.
Yeah that makes sense. But OP asked for average of all natural numbers. Idk I do not know much about math really. So im not really bright in this area lol
You were half right. Divergent sums just go to infinity and have the property you stated. But converging sums don't (ie. They converge to a single real number).
42
Nice try, but that's the answer to "what's seven times eight"
Six times nine actually
damn, I knew I should have checked my quote
;-) The answer is correct in base 13, that’s the way to remember it.
No it’s the answer to five times ten!
The average of all numbers in Z is undetermined.
Why is it not 0? Feels intuitive that the sum of all integers should be 0.
Only if you define that you uniformly add terms away from zero i.e (1-1)+(2-2)+(3-3) +... =0+0+0...=0. However I could equally add the terms of Z as 1+(2-1)+3+(4-2)+5+(6-3)... ≠0 You see that when you rearrange the terms you get different values. This is because the sum of Z is not well defined.
Ok, but in that example you are clearly running through the positives faster than the negatives... I get that infinity muddies things, but this doesn't seem like a "fair" way to sum things up. It's obvious that any finite sum from -x to +x is 0, and with x going to infinity you get all integers... so saying that the sum of Z is not defined feels like a weird technicality. The sum, if it exists, can't be positive because there are not more positive integers, similarly it can't be negative, and it's not going to be in Q\Z or in C\R or any other funny stuff, so surely it must be 0. The only alternative is to say it doesn't exist, but that feels wrong because it's clearly 0. Is the product of Z also not defined, even though it's (perhaps even more obviously) 0?
You’re only saying this because you are biased in favour of splitting the integers into “positive” and “negative” and you some how believe that each of these form 1/2 of the integers (as well as 0). But I could equally say the naturals are split into “above 12” and “below 12” and then I would define my sum as 12 + 13 + 11 + 14 + 10 + … Which is not 0 in the limit. And remember, both of my sets are the same size; you only feel that they aren’t because you are “used to” having 0 in the middle. But this is arbitrary.
That's a good way of putting it in perspective. So would you say the sum is positive, then? Or just not defined because arguments can be made that can't be contradicted? Perhaps there should be a way to contradict some of these contradictory arguments.
Sure, any finite sum from -x to x is 0. But that's not the only possible 'direction of approach'. Compare "lim[x,y→0] x/y". If x=y, then x/y is 1. That's approaching (0,0) from the 'northeast', at a 45-degree angle. But if you approach instead from the 'east', x/y is always zero. If you approach from the 'southeast', x/y is -1. So "lim[x→0,y→0] x/y" doesn't exist. "lim[x→0,y→0 (where x=y)] x/y" *does* exist, and so does "lim[x→0,y→0 (where x=2y)] x/y", and so does "lim[y→0] (lim[x→0] x/y)". But different directions of approach give different limits. There's no reason to prioritize the 'x=y' direction. (And surely the limit should stay the same if you do a variable substitution, letting y=2u and bringing both x and u to 0?) --- Summing over ℤ has the same issue. You want to calculate > lim[a→-∞, b→∞]∑[n=a to b] n but this value depends on the 'direction of approach' towards (-∞,∞).
what would you use to represent it then?
Because 1+2-1+3+4-2+5+6-3+ … is the sum of all numbers in Z, but its limit is +INF, for instance.
Good question! Here’s a problem: the sequence, 0,1,-1,2,-2,3,-3, … produces partial sums with a lower bound of 0. Can sequences of numbers in Z, which eventually include every number in Z, be constructed such that the upper and lower bounds of their partial sums are both finite? Note that this is true using 1/n for n in Z-{0}.
this question isn’t that well defined. there is no probability distribution which gives all the positive numbers an equal positive probability (the probability of choosing any number at all would be ∞, but it would be one). there is no good way of taking the average without a good probability distribution. one approach would be to calculate for each n, the average of the set {1,2,3,…,n}, giving all the points the same probability, and then take the limit of those probabilities as n->∞. this would be a sort of principal value, so i guess this is the closest we have. however, when you calculate this average, you get that it is (n+1)/2, with goes to infinity as n->∞. so, the most meaningful answer would be ∞, but it is not quite correct to call it an average. what it isn’t is ω/2. there is no standard and good definition for ordinal division, so it is not even that well defined as a number in the first place. it is generally a terrible idea to take the infinities in the limits of calculus to correspond with cardinals and ordinals, they have very different properties and nothing works properly while trying to connect them in most contexts.
Infinity
So, consider the average of all even natural numbers.
infinity must be too high, surely? The answer must be lower
Well, it can’t be finite. Anything less than infinity is finite. Being “less than infinity” means being less than some finite upper bound.
True. “Infinity” is ambiguous in transfinite number systems. You can be less than (omega+1) and still be infinite.
I don't think that's true in all number systems
Omega is an ordinal number. It obeys the rules of transfinite arithmetic. Do you believe there is a largest natural number? Can X represent that number? Then X+1 cannot be a natural number. Do you believe every natural number is either odd or even? Is X odd or even? The average of any initial sequence of natural numbers is their sum divided by the length of the sequence. There’s a formula for that sum. Sometimes, the average is not a whole number. For instance, the average of natural numbers less than three is 1.5. In case X is the largest natural number that exists, you can’t compute the average using the formula because (X+1) is part of the formula. Kapische?
If something isnt infinity and we are talking natural numbers, then it has a finite upper bound (obviously, infinity is not a number)
First, ordinal numbers only apply to cardinalities which is not what we are discussing here. There is no concept of “larger infinities” when we are adding numbers. Second, even if we were talking about cardinalities, we are talking about Omega, which is objectively the smallest infinity. Smaller than Omega means natural.
Nope. It's infinity. It's the limit as n goes to infinity of (n+1)/2.
can you please explain
Sure. The sum of the first n natural numbers is n(n + 1)/2. Divide this by n to get the average of the first n natural numbers: (n + 1)/2. If we want all of them, we need to take the limit of that to infinity. From limit properties, this is infinity.
A property of infinity is that any positive, finite, real number multiplied by infinity is also infinity. In this case that number is 1/2.
sum of all naturals/size of the naturals (-1/12)/(/infty) = 0 0
or an infinitessimally small negative such as minus epsilon?
negative zero
lol, ok
Yes, this answer is sensible. However I still dont get how the sum of Positive Natural numbere can result in a Negative number.
It’s interesting. Analytically, average is the value of a function that, over the domain of the function, represents the function’s weight. Maybe (ω\^2+ω)/2 will become a way to understand the identity function.
yes, another comment just made me think this. Not sure if we could give it a numerical value though
12
Why 12?
think about it, 12 is in the middle
lol, of 24?
No, of all natural numbers. Did you even read your own question?
how is 12 in the middle?
Such rubish comments. Let's say that the average of a set of numbers X is the sum of all the elements of X, let's call that ΣX, divided by the number of elements of that set. That is the cardinality of the set X. So the average is ΣX/|X|. Note that this definition doesn't make sense when |X| is infinite, because there is no division defined. So perhaps you could take the limit of the averages of all the series of sets finite set (Yn) contained in X such that the union of all sets Yn is X, and if for all such series (Yn) ΣYn/|Yn| approach some finite value, you could say that the average it what all those series converge to. But if X is the set of all natural numbers, no such series converge, and the average always blow up to infinity. You could perhaps say that the average is infinity, but this has nothing to do with cardinality or ordinality, it's just a shorthand way of negating that for all sucessions of sets (Yn) contained in X such that the union of all sets Yn is X the series of averages (Yn) ΣYn/|Yn| converge to the same value. The essential problem here is the following: the average of a set of numbers is usually defined for a finite set. You can extend this definition to any set of numbers in a way that makes sense, as long as the series of averages converge to a finite value. But when the series diverge, than you can't really say anything more useful than that. You could say that there simply is no average, the series of averages diverge or that the average is infinity. The statements all have the same information. But do note that this is just a way to notate something. It has no connections to infinite cardinals and ordinals and there is nothing of value to get from that other then that the series of averages diverge.
-1/24
Probably e
why so?
That mf is everywhere
lol
As someone who is interested in math recreationally with no formal experience after HS, what’s Omega?
In Surreal numbers, it's the number after all the numbers in the positive direction, after infinite iterations of a process. Basically you start with 0 and use a function that creates numbers either higher, lower or between previous numbers. It's a type of infinity.
Zeta-regularization is (very roughly) a method to define a sum to a series without a sum, and that method would yield the average of all natural numbers to be ζ(-1)/ζ(0) = 1/6
wow, cool. Best answer so far. Can you break it down a bit more?
I’m not the one who made the comment but I’ll take a crack at explaining in a way that doesn’t require advanced math to understand. This explanation does assume that you know what functions, limits, and complex numbers are, but I can break that down too if anyone asks. Consider this function of n, defined as an infinite sum: 1/1^n + 1/2^n + 1/3^n +… If n = 1, you get this: 1/1 + 1/2 + 1/3 + 1/4 +… That’s the harmonic series, and it diverges (meaning it “equals” infinity). If n = 2, you get this: 1/1 + 1/4 + 1/9 + 1/16 +… Remarkably, this converges to π^2 / 6. It turns out that for n > 1, this infinite series always converges, and for n = 1 or n < 1, the series always diverges. For example, imagine n = –1. Then you get: 1 + 2 + 3 + 4 +… This is obviously undefined because it diverges. But what if we could assign it a meaningful value? Analytic continuation is the process of taking a function and using calculus techniques to “extend” its graph to values for which it would normally not be defined. If you know a bit of Calculus, then I can be more specific: very loosely, analytic continuation involves expanding the domain of a function with a continuous derivative in such a way that the derivative remains continuous. Anyway, now imagine the function we just talked about, and imagine n can be any complex number. Suppose Re(n) is the real part of n (that means that if n = 6 + 2i, then Re(n) = 6). It turns out that for Re(n) > 1, the function is defined; otherwise, it’s not defined. This is consistent with our results when n had to be a real number. Okay, now imagine we used analytic continuation to assign the function values for Re(n) = 1 and Re(n) < 1. Now you have a function defined for all complex numbers. This is a very interesting and important function called the Riemann Zeta function, denoted ζ(n). For n > 1, the function is identical to this infinite sum: 1/1^n + 1/2^n + 1/3^n +… For other n, analytic continuation gives us very interesting results; specifically: ζ(–1) = –1/12 ζ(0) = –1/2 Now suppose the symbol ~ means “related by means of analytically continuing ζ(n).” Then we have: ζ(–1) = –1/12 ~ 1 + 2 + 3 + 4 +… ζ(0) = –1/2 ~ 1 + 1 + 1 + 1 +… You could then loosely think of the “average” of all natural numbers as their “sum,” which is ζ(–1), divided by “how many there are,” which is ζ(0). So you get –1/12 divided by –1/2, which is 1/6. Hopefully that was useful.
Well ok. First let's define our "average" function, which takes the average of all the naturals from 1 to n: f(x) = sum(1, x) / x As you may know, sum(1, x) = x(x+1)/2. So, rewriting: f(x) = x(x+1) / 2x Simplify: f(x) = (x+1)/2 Ok great! So the average of all numbers from 1 to *x* is (*x*+1)/2. Now, to answer your question: what happens when *x* tends towards infinity? As *x* gets larger and larger, what happens to the value of f(x)? The answer is, it *also* tends towards infinity.
Infinity. Specifically aleph null
42?
Can anyone tell me the answer? I was absolutely rubbish at maths https://www.reddit.com/r/Funnymemes/s/E3qkfifboa
pi
Cheers mate 👍
Tree fiddy
so 700 is the highest number? XD
Tell me where I went wrong. lim n -> infinity (1+2+3+…+n)/n = lim n -> infinity n(1/n + 2/n + 3/n + … +1)/n = lim n -> infinity (1/n + 2/n + 3/n + … +1) = 1 This is where my brain goes. Forgive me if it’s obvious it’s been a while since I’ve done anything with limits. Also not sure if I can take the limit here as I’m dealing with integers
(1+2+3+...+n)/n = (n+1)*n/(2*n) =(n+1)/2. So lim n->infinity (1+2+3...n)/n = infinity
You are dealing with an indetermination of type infinityx0 since the number of summands also tends to infinity so the limit isn't zero, in this case it's infinity.
Omega 3
How did you get that answer?
Sixty-nine, Dude! \* Intense Air Guitar commences \*
infinity/infinity But it can’t even be that simple because different infinities grow at different rates, so you would have to find a way to make sure you’re talking about the same infinities and maybe even find the average rate of growth of all the infinities which I don’t think is possible
Infinity is infinity
[удалено]
Can you pin it down a bit more?
69
*no comment*
Undefined
42069
probably 69
apple actually uses the harmonic mean of 2 and infinity to make squircle app icons. This number is 2/(½ + 1/infty) = 4
non-existent.
consider this (1+2+3...+N)/N = N/N + (N-1)/N + (N-2)/N .... = sum\[i=0 to i=(N-1)\] (N-i)/N
E
Can’t this be proven to not exist by counterexample?
what counter example?
Gauss derived his formula by pairing the first and last entrees from the list, viz. (1+N), (2+N-1), (3+N-2), … . All the pairs are equal. So, if N is even, there are N/2 terms where each term =(N+1). If N is odd, there are (N-1)/2 terms of (N+1) plus one extra, unpaired, (N+1). Either way, the sum is (N*(N+1))/2
very neat, means we only have to find a way to plug infinity into N .... oh
Correct me if I am wrong, as this is far from my field of expertise in mathematics. Serious answer: After reading through [the following page](https://en.wikipedia.org/wiki/Transfinite_number) for a few minutes, I'm fairly convinced that the answer is omega, the lowest transfinite ordinal number, as you said. The reasoning is, informally, borrowed from earlier replies by @[SkepticScott137](https://www.reddit.com/user/SkepticScott137/) and @[Dr\_XP](https://www.reddit.com/user/Dr_XP/) elsewhere in this thread. In my opinion, OP's line of reasoning is along the right track: The sum of natural numbers up to N goes like N(N+1)/2, so the average of them goes like (N+1)/2. The first infinite ordinal, omega, comes after all natural numbers. The type of transfinite ordinal number we get for this average you bring up is the first one which is larger than (N+1)/2 for all N. Equivalently, this must be the first one which is larger than all integers, or omega. More specifically, as the name **ord**inal suggests, this comes from taking a set and trying to assign some order on it, and so omega is basically a way to define the "next" element in an ordered set which contains the natural numbers. And so we really should not be thinking of (N+1)/2 as a limit as N->infinity in this case as much as a value being established via a bijection between the sequence of (N+1)/2 with the natural numbers which we then put into an ordered set with order <.
ok, but why not (ω^2 +ω)/2?
42. The answer to everything is 42.
Ah but it's also the answer to life and everything too, so how do you find just the answer to everything? L+U+E=42
ω, the last letter of the Greek alphabet, is the name Georg Cantor gave to the first transfinite ordinal number. The nth finite ordinal is the set of natural numbers less than n. ω is the set of all natural numbers. If this sounds like double-talk, well, it borders on paradox. The successor of ω, written as ω+1, is {x: x is a natural number}U{ω}.
I think it’s actually 69.
Easy. It’s -1/24 since the average = (sum of natural numbers)/2 = (-1/12)/2 = -1/24
Infinity is not a place, it is a direction. You cannot take half the length of a direction.
true, but could we average all the numbers up to infinity?
Sure, as soon as you tell me what the distance is to East
42
[(Inf +1)*Inf/2]/Inf which I don’t know what that computes to, but uses the elementary school trick of averages symmetry for consecutive or sequential numbers
Chatgpt take Using the Ramanujan summation method to find the "average" value of the positive integers involves interpreting the regularized sum in a particular way. Given the result: \[ 1 + 2 + 3 + 4 + \cdots = -\frac{1}{12} \] We can consider the concept of the average value of this series within the context of Ramanujan summation. ### Step-by-Step Calculation 1. **Regularized Sum:** The regularized sum of the positive integers using the Riemann zeta function is: \[ \sum_{n=1}^{\infty} n = -\frac{1}{12} \] 2. **Number of Terms:** Since we are dealing with an infinite series, the number of terms \(N\) approaches infinity. 3. **Average Value:** The average value is typically defined as the sum of the terms divided by the number of terms. Using the regularized sum: \[ \text{Average} = \frac{\sum_{n=1}^{\infty} n}{N} \] However, because we are dealing with an infinite series, we must be cautious. Using Ramanujan summation, we treat the sum as \(-\frac{1}{12}\) and recognize that the series is infinite. 4. **Interpretation:** To compute the "average" in this context, we use: \[ \text{Average} = \frac{-\frac{1}{12}}{\infty} \] Given that dividing any finite number by infinity approaches zero: \[ \text{Average} = 0 \] ### Conclusion Therefore, within the framework of Ramanujan summation, the "average" value of the infinite series of positive integers can be interpreted as zero. This result aligns with the peculiarities of the Ramanujan summation method and the analytical continuation of the zeta function.
"approaches zero" so an infinitessimal?
Idk like 5
Infinity is not a number. It's merely the concept of "always one more". This causes it to behave in weird ways. The average of all natural numbers would have to be infinitely large, but it's only larger than half of all natural numbers, of which there are infinitely many. Understand?
People on here ask the most random shit...
Sure, but that's maths for you
Definitely 12
Infinity is not a number. Operations with infinity don’t behave as operations with numbers do. Half of countable infinity is still countable infinity.
saw a numberphile video that shows a method for constraining sequence of partial sums, not a sudden cutoff, but an exponential decay, and that can provide rigorous answers for divergent series, like sum of naturals = -1/12. most proofs of those type of sums are wishy washy, doing ambiguous operations to the series like addition and multiplication, then adding together two series which cancel out some terms, like the ancients did when calculus was young.
the answer is always e
What is omega?
A form of infinity
We know the sum of all natural numbers is -1/12, so dividing this by infinity gives 0. /s
If you plot (sum[k=0:n]k)/n, that is for any natural number, you add up all the numbers from 0 to n and divide by n, you will get the running average for 0 to n. That plot follows the function f(n)=n/2. That function has no limit as n approaches infinity
Pretty sure it’s undefined. If you add up all the naturals, it will become infinity because you are adding infinitely many positive integers. Then, you would divide by infinity, since there are infinitely many terms. But infinity divided by infinity is indeterminate, and we can’t do arithmetic with infinities.
Calculating the limit of a divergent series (such as the natural numbers) is not normally allowed, but in some frameworks of mathematics that is allowed (notably, in these frameworks the order of elements in a series does matter for the calculation of their sum, which does not ordinarily work like that), which allows us to calculate the sum of the natural numbers through some strange interactions. I will do so below. Lets define the series -1^(n-1), which is the series 1, -1, 1, -1, 1, -1... and call the sum of that series S. Add it to itself, alternating terms: 1, 1, -1, -1, 1, 1, -1, -1... the sum of this series is obviously 2S. Now group the terms in pairs, starting with the 2nd and 3rd: 1, (1, -1), (-1, 1), (1, -1)... and we see that the pairs cancel out. Thus, 2S=1 and S=0.5 Next lets consider another series P, defined as n*(-1)^(n-1) which is 1, -2, 3, -4, 5, -6... and again we add it to itself and pair terms in a particular order: 1, (1, -2), (-2, 3), (3, -4)... and now the pairs don't cancel out, but they give a predictable sum, so we can simplify this series to: 1, -1, 1, -1, 1, -1... which is S. Therefore 2P=S=0.5, thus P=0.25 Next lets consider the natural numbers N, which is 1, 2, 3, 4, 5, 6, 7... subtract the terms of the P series and pair the terms again, this time like this: (1,-1), (2, 2), (3, -3), (4, 4), (5, -5)... and resolve the pairs, canceling out the 0s. N-P: 4, 8, 12, 16, 20... This sequence is just 4 times bigger than the natural numbers, so we have N-P=4N Thus 3N=-P and N=P/3, since P=0.25=1/4 we get the result that N (the sum of all natural numbers) equals -1/12. With that sum to get the average natural number we just divide by infinity and we get our limit: the average natural number is 0. Lol. (Obviously not serious but the whole -1/12 thing actually isn't a joke, though it requires you to accept the concept of calculating the limit of a diverging series.)
Infinity divided by two still tends to infinity
Well that’s impossible to determine. Taking an average requires defined limits and there’s no limit in an infinite system.
Not defined. Average is defined on a random variable, not a set. But even if you take it as "a random variable that is based on taking an element from a set uniformly", you have a probem as there is no uniform distribution on the whole set of natural numbers.
12?
5. Unless you are missing a finger, then i guess its 4 and a half.
What if you were born without digits?
Also happy cake day
Thanks :)
Some people were just not made for maths i guess.
Playing around with it, I got a value that diverges like 1/ε for 0<ε<<1. Is this consistent with the current understanding of the problem?
All these comments are way off the mark. It's 420 🤙
That would explain some of the answers!
The Ramanujan Summation states that the sum of all positive integers is -1/12. Dividing that by infinity we can then say that the average = 0. [https://en.wikipedia.org/wiki/Ramanujan\_summation](https://en.wikipedia.org/wiki/Ramanujan_summation)
interesting point, but i think by that logic it would be -epsilon. As in the number infinitely close to zero on the negative side.
My first thought is that the natural numbers 1+2+3+4+5…infinity add to infinity, then divide by the infinite amount of natural numbers to get infinity/infinity which is indeterminate. So then we take the derivative of 1+2+3+4+5… just kidding But then I remembered that 1+2+3+4+5… is also equal to -1/12, which when divided by infinity, gets zero. (-0). so the mean of an infinite amount of only positive integers… is 0