the sum of all natural numbers is negative one-twelfth?

Posted: November 16, 2012 in Mathematics
Tags: , , ,

No, I haven’t gone crazy yet.  And yes, I am telling you that the sum of all natural numbers (aka positive integers) is -1/12.  This maybe a little bit hard to swallow because what I am saying is that if you add up all the counting numbers (ie. 1+2+3+…), you will get a negative number and it’s finite and a “fraction” too.  This sounds crazy I know but I can show you how it could be done, so read on.  First of all let me give you some concepts on summations.

Cesaro Summability

Cesaro summation could be used when one wants to assign a meaning to the sum of a divergent series.  In simple English, if an average of the nth partial sum of a series converges to a limit, then that limit is the Cesaro sum of that series.

If \displaystyle\lim_{n\to\infty} \frac{s_1+s_2+\ldots+s_n}{n}=L exists, then L = Cesaro sum.

Now let’s examine the following alternating sequence,


The sequence of the partial sum is,


Then a sequence of \displaystyle\left\{\frac{s_1+s_2+\ldots+s_n}{n}\right\} is,


Hence the limit,

\displaystyle \lim_{n\to\infty}\frac{s_1+s_2+\ldots+s_n}{n} = \frac{1}{2}.

If you think this is too much math and too difficult to understand, I can give you a naive way of obtaining the same result of that alternating series.  Let’s ignore the rigorous part of summability here at the moment, and be open minded, assuming a divergent series can be rearranged.

S = 1-1+1-1+\ldots = 1-(1-1+1-\ldots) = 1-S,


\displaystyle S=1-1+1-1+\ldots=\frac{1}{2}.

If you believe this is true, then let small s be the sum of all natural numbers,


s-4s = -3s = 1+2+3+4+5+6+\ldots -4-8-12-\ldots

-3s = 1-2+3-4+5-6+\ldots

-3s= 1- (2-3+4-5+6-\ldots)

-3s= 1- (1-2+3-4+5-\ldots + 1-1+1-1+1-\ldots)

-3s= 1- (1-2+3-4+5-\ldots) - (1-1+1-1+1-\ldots)

Since the sum of that alternating series is \displaystyle\frac{1}{2},

\displaystyle -3s=1+3s-\frac{1}{2}

\displaystyle \therefore s = -\frac{1}{12} 😀


This result is somewhat related to Riemann Zeta Function,


If we try to compute \zeta(-1), we get the following result.

\displaystyle \zeta(-1) = \sum_{n=1}^\infty \frac{1}{n^{-1}} = \sum_{n=1}^\infty n=1+2+3+\ldots,

which is the sum of all natural numbers.  Now if you want to compute \zeta(-1), you can click the link here.

Wolframalpha tells us \displaystyle \zeta(-1) = -\frac{1}{12}.

  1. John Allsup says:

    The wikipedia page has a few good points. The sum notation for the zeta function zeta(s)=sum 1/n^s is only valid for s>1. The zeta(-1)=-1/12 is arrived at via analytic continuation, but the sum notation is not valid here. It is interesting that the ‘hocus pocus’ approach using 1-1+1-1… arrives at the same result as the zeta analytic continuation, though.

    • Jonas Kölker says:

      Note that sum(1, …, n) = n(n+1) / 2, and so average(1, …, n) = (n(n+1) / 2) / n = (n+1) / 2. This is divergent. There is no N such that for all n >= N we have (n+1) / 2 < 0—there isn't even a single n for which that's true—so the prefix average never gets within 1/12 of (-1/12).

      I think your first shennanigans are in saying that "-3s = 1 – 2 + 3 – 4 + 5 …". It seems to suggest that -3s is a series with elements +1, -2, +3, …, which is not the case: its elements are -3, -6, -9, and so forth. It *may* (at least a priori) be the case that -3s and not-really-minus-3s [s_n = n * (-1)^(n+1)] both have a Cesaro sum and that the two sums are equal, but those three statements would have to be proved separately. (Given that I have disproven one of these statements, this can't be done).

      I think the biggest lesson to take away from your post is this: when dealing with infinite series, moving an element from one index to another is forbidden! Or rather, the result is a different series. Likewise for arbitrarily deleting or inserting elements (in particular zeroes). I think I can prove that if you only swap finitely many elements of a convergent series, you get another convergent series with the same limit, so the not-quite-but-similar-to-commutative-law holds in the finite.

      Another way of getting at the flaw in your "proof" is to recall that sequences over some set X (typically the reals) are functions from the natural numbers to X. The flaw would be a lot harder to hide if for each line of your calculations you added definitions on the form [s_n = …] for each of the sequences you're working with. Especially if you tried to show how s1_n + s2_n = s3_n etc. for every combination of sequences you use.

      Sorry for 1. stating the obvious; 2. ruining your joke by explaining it; and/or 3. spoiling an intriguing puzzle. But the truth must out 😉

      • Problem is you are using partial sums, which not the same thing as infinity as a whole. Using partial sums is fine for convergent series, but is not valid for divergent series.

    • 1-1+1-1+1-… is actually the limit xi(0)=-zeta(0)=1/2 which is why this value is accepted.

  2. […] The sum of all numbers is negative one-twelfth?, Mathgarage […]

  3. Neil Bates says:

    I have found a direct contradiction in the treating of the infinite sum of positive integers (1 + 2 + 3 + 4 + 5 + …) = -1/12. Yes I know, it can be related to that number by the zeta function etc. but the concept is not algebraically consistent. I don’t just mean regarding rearranging associative parentheses either. The infamous proof in the circulating Youtube makes use of offset series additions to support the idea that 1 – 1 + 1 – 1 + 1 – … = 1/2 and that 1 – 2 + 3 – 4 + 5 – … = 1/4. Combining those with another offset, there is a “consistency” argument that 1 + 2 + 3 + 4 + 5 + … = -1/12. Actually I think it’s a clever paradox that apparent “algebra” supports the apparently unrelated result from higher math. However, there is a true contradiction as I show below (hoping for decent formatting but write it out as needed):

    1 + 2 + 3 + 4 + 5 + … “= -1/12”
    …..-2 – 4 – 6 – 8 – … “= 1/6”
    1……….-1 – 2 – 3 – …. = 1 1/12 OR 1/12?

    Sliding the -2x of the original series over more and more creates ever larger discrepancies. It is inconsistent.

  4. dan says:

    I think we can explain the result with conventional math. Isn’t the expression s-3s halfway down the page equivalent to saying infinity minus infinity? I am not a mathematician, but I believe that that expression equals any number you want, including -1/12

  5. Wang Kerr says:

    Hmmm what worries me is that if I take a supercomputer and do the sum it will diverge pretty quick and certainly will never be negative.

  6. Asad Mirza says:

    That’s just it no supercomputer could add up an infinite amount of natural numbers, nothing can, which is why you have to use these methods to get to the answer.

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