No, I'm not talking about $-\frac{1}{12}$.

I was talking with someone the other day, and they said that the sum of all integers, positive and negative, is zero because they all cancel each other out. Basically,


I'm skeptical, given how infinity tends to never work the way I want it to, but I have no math to back this.

What I'm asking is this:

Is the sum of all integers 0?

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    $\begingroup$ Tbh it looks like Zero to me. -1/12 + 1/12 = 0 $\endgroup$
    – Neil
    Commented Mar 29, 2015 at 1:06
  • $\begingroup$ I agree with Neil, the only sensible sum is zero for the same reason he mentioned. $\endgroup$
    – user541686
    Commented Mar 29, 2015 at 2:53
  • $\begingroup$ $$\lim_{N\to\infty}(-N)+(-(N+1))+(-(N+2))+\dotsb+(N-2)+(N-1)+N=0$$ That's basically the only way to deal with this matter. $\endgroup$ Commented Mar 29, 2015 at 3:46
  • $\begingroup$ As explained in the answers, there are several methods of summation. This particular method corresponds to the Cauchy principal value. $\endgroup$
    – Siméon
    Commented Jul 26, 2015 at 0:09
  • $\begingroup$ The sum does not make sense. $\endgroup$ Commented Jul 1 at 4:36

6 Answers 6


We model the concept of "infinite sum" by taking limits of finite sums. This model has the "flaw" that the value of an infinite sum will depend on the order we sum the terms (compared to finite sums, where the order doesn't matter). If you think the sum of all integers as

$$\lim_{n\to \infty} \sum_{i=0}^n (i-i)$$

then certainly the sum is equal to $0$, however, if you think this sum as

$$\lim_{n\to \infty} \left(\sum_{i=0}^n (i-(i+1))\right)$$

(All the integers appear as a sumand) then the sum is not zero!, The limit doesn't exist.

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    $\begingroup$ Thanks, this answer makes the most sense to me. $\endgroup$ Commented Mar 28, 2015 at 20:47
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    $\begingroup$ It might be worth pointing out that the first limit only exists due to grouping two integers together each time. Taking $n$ and $-n$ separately as successive terms of the sequence gives a divergent series $\sum_{i=0}^\infty (-1)^i\lfloor i/2\rfloor$. $\endgroup$ Commented Mar 29, 2015 at 4:19
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    $\begingroup$ Nice. Other answers mention reordering the terms, but this answer shows that we can get different sums merely by changing the association (grouping) of terms without changing their order. $\endgroup$
    – PM 2Ring
    Commented Mar 30, 2015 at 10:12

Not really. It's not even a particularly well-posed question, to be honest. The problem is that most the usual suspects behind assigning values to divergent sums (like the statement that $1+2+3+\ldots=\frac{-1}{12}$) work on sums of the form: $$\sum_{i=1}^{\infty}s_i$$ for some sequence $s_i$. Why is this a problem? Well, they simply can't handle it if your series extends infinitely in both directions.

Okay, we could try to work around it by asking about $$0+1-1+2-2+3-3+4-4+\ldots$$ though I'm not sure what techniques might handle such a sum. However, the trouble is that we could equally well ask about: $$0+1+2-1+3+4-2+5+6-3+7+8-4+\ldots$$ where we include two positive terms for each negative one - and we'll probably get a different answer. So what order are we supposed to choose?

To illustrate the point, we can use the usual pseudo-algebraic techniques, or something more rigorous (like a Cesaro summation) to assign the value $\frac{1}2$ to the sum: $$0+1+0-1+0+1+0-1+0+1+0-1+0+1+0-1+\ldots$$ but, if we shuffle the zeros around to get the expression: $$1-1+0+0+1-1+0+0+1-1+0+0+1-1+\ldots$$ we end up assigning a different sum ($\frac{1}4$), despite the terms being the same. So we have some pathological behavior based on the ordering - which is troubling, because if we have no canonical order (i.e. we want to sum the set $\mathbb Z$, not some enumeration of it), we have no reason to think we can't order it so as to convince our method to spit out whatever number we want!

Sadly, the notion of summing a set is, almost exclusively, a measure theoretic concept applying to only absolutely convergent series - it doesn't even work on conditionally convergent series, let alone divergent ones. Worse, other techniques tend to want to examine a generating function, but they're not bidirectional unless we start talking about coefficients of $x^{-1}$ and so on (which we could do, but wouldn't be as useful as usual, since the main advantage of such a generating function is being able to work with the series near $0$, but we just plopped a singularity there...)

  • 3
    $\begingroup$ +1, especially for being the only answer to use the word "absolutely convergent". $\endgroup$ Commented Mar 29, 2015 at 18:35
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    $\begingroup$ +1, but it looks like you forgot to write your footnote :-) $\endgroup$
    – k_g
    Commented Mar 29, 2015 at 21:04

Certainly $\sum\limits_{n = -\infty}^\infty n$ is divergent because for each positive integer $N$, $\sum\limits_{n = -2N}^{2N+1} n = 2N+1 \ge 2$.

It appears that the person you spoke with interpreted $\sum\limits_{n = -\infty}^\infty n$ as $\lim\limits_{n\to \infty} \sum\limits_{i = -n}^n i$, in which case the sum will be zero. However, this is incorrect. A bi-infinite series $\sum\limits_{n = -\infty}^\infty a_n$ converges if the double sequence

$$A_{m,n} := \sum\limits_{i = -m}^n a_i$$

converges. Note $m$ is independent of $n$. When $a_i = i$, we have $A_{2N,2N+1} \ge 2$ for all $N\in \Bbb N$, by the statement made above. This implies that the double sequence $A_{m,n}$ is unbounded, hence divergent. So $\sum\limits_{n = -\infty}^\infty n$ is divergent.

However, if you know a priori that a bi-infinite series $\sum\limits_{n = -\infty}^\infty a_n$ converges, then you can claim that the sum of the series is $\lim\limits_{n\to \infty}\sum\limits_{i = -n}^n a_i$.


When you see written $\sum\limits_{i=1}^{\infty} x_i$ this term has a sense only if the series of $x_i$ is converging of diverging towards $\infty$ or $-\infty$. If you write $$ ... + (-3) + (-2) + (-1) + 0 + 1 + ... = 0$$

this sum is actually not defined. What would be the term of your sum? You could write $\sum\limits_{i=1}^{\infty} i$ + $\sum\limits_{i=1}^{\infty} -i$ which gives you $\infty - \infty$ but this is not defined.


EDIT: Shoot. I just realized that the op wasn't asking about the -1/12. I can't figure out how to withdraw the answer. Oh well, what I posted is still relevant to the sums of infinite series.

The sum is undefined, as other respondents have noted. If I recall correctly, this is an example of what's called "Ramanajan Summation." There's a wiki article on that subject. I'm new to supplying answers, so I don't know if linking in from wikipedia is kosher.

Anyhoo, the Ramanajan sum is supposed to be a way of giving a value to an infinite sum in order to study the properties of the partial sums, but isn't (I think) meant to be taken too seriously as the sum of the entire series.

Technically, you're supposed to put a little notation after the sum to indicate that you're not using a regular sum. E.g.:

1+2+3+4+... = 1/12 (R)

where the "R" denotes Ramanajan summation.

Hope that helps...

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    $\begingroup$ To your comment at the start of the edit: There should be a delete button visible to you. Though, I'm not sure given that you're an unregistered user. (You could always flag and ask a moderator to delete it if you really want it gone) $\endgroup$ Commented Mar 29, 2015 at 3:18

It is generally considered that the series $$ 1 - 1 + 1 - 1 + 1 - 1 \pm \cdots$$ does not converge, since the partial sums alternate between $1$ and $0$ forever.

Even if you suppose that it's OK to rewrite a double-ended sum by taking terms alternately in each direction, you do not get a better result by summing

$$ 0 + 1 - 1 + 2 - 2 + 3 - 3 + 4 - 4 \pm \cdots.$$

You could put parentheses around pairs of terms and pretend that both of these series are the same as $ 0 + 0 + 0 + \cdots,$ but they are not.

You can try rearranging the positive and negative integers in any sequence you like; none of it is going to stop the series from eventually taking larger and larger steps up or down with each new partial sum.


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