# Why is $1+2+3+\cdots = 0$? [duplicate]

I had seen this result a while back in a Numberphile video:

$1+2+3+\cdots = -\frac{1}{12}$

I was trying to prove the same result using a different method when I accidently proved that the sum was 0!!

I am unable to find out what mistake I have made. Or is it that perhaps the sum is not really $-\frac{1}{12}$? Perhaps it is something paradoxical?

### Here is the proof:

\begin{align*} S= & 1+2+3+4+5+...\\ = & (1+3+5+7+9...)+(2+4+6+8+...)\\ = & (1+3+5+7+9...)+2(1+2+3+4+...)\\ \implies S= & (1+3+5+7+9+...)+2S\\ \implies -S= & (1+3+5+7+9+...)\\ \\ \end{align*} Now, we subtract S from the above obtained -S as follows: \begin{align*} -S= & 1+3+5+7+9+...\\ S= & 1+2+3+4+5+...\\ ---- & ---\text{subtract}-------\\ -2S= & 0+1+2+3+4+...\\ \implies-2S= & S\\ \implies-3S= & 0\\ \implies S= & 0 \end{align*}

## marked as duplicate by Jack D'Aurizio sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 28 '15 at 13:22

• Woahhh......waiting excitedly for an explanation for this – The Artist Aug 28 '15 at 9:53
• As explained multiple times on the site, all this assumes that S is a finite real number (otherwise what is the meaning of substracting S?). – Did Aug 28 '15 at 10:04
• Because the sums your are manipulating are not convergent in an usual sense, the above manipulations like rearranging terms are not sensible and yield nonsense results. You can produce arbitrary outcomes by different rearrangement/addition schemes... math.stackexchange.com/questions/39802/… – tired Aug 28 '15 at 10:05
• @shivams: Just because someone else did it doesn't make it right. – Asaf Karagila Aug 28 '15 at 10:14
• @shivams Exactly the reason why the infamous video you are referring to did a wonderful job at making its authors look as wise asses and a terrible one at explaining maths. – Did Aug 28 '15 at 10:18

The result $1+2+\cdots=-\frac{1}{12}$ is an example of using a summability method for assigning finite values to divergent series. It should be instructive to review what this actually means.

An infinite series $\sum_{k=1}^\infty a_k$ is a special case of a limit $\lim_{n\to\infty}\sum_{k=1}^na_k$, specifically a limit of partial sums. If we consider the vector space $V$ of all convergent sequences $(s_1,s_2,\cdots)$ then $\lim_{n\to\infty}$ is a linear map from $V$ to the scalar field (most often $\Bbb C$). If $W$ is any larger vector space of sequences which contains $V$ but also contains nonconvergent sequences, and if we extend the map $\lim:V\to\Bbb C$ to a linear map on the larger domain $L:W\to\Bbb C$, we can use $L$ to assign "limits" to nonconvergent sequences, or at least those in $W$. Thus, by identifying an infinite series $\sum_{k=1}^\infty a_k$ with its sequence of partial sums $s_n:=\sum_{k=1}^n a_k$, we can assign finite values to some kinds of divergent series.

Many manipulations of convergent series which preserve their limit no longer preserve the finite value assigned by $L$ to divergent series. In particular, while it's still okay to swap any finite number of terms in the sequence of terms, and since $L$ is linear it is still okay to scalar-multiply every term in the sequence and it is okay to add series term-wise, it is not okay to shift terms and to insert or delete $0$s between terms. This may change the values of divergent series!

The summability method at play here is zeta regularization. Given an infinite series $\sum_{k=1}^\infty a_k$, which remember we identify with the sequence of partial sums, to perform zeta regularization we define the zeta function $\zeta_A(s)$ by the Dirichlet series $\zeta_A(s):=\sum_{k=1}^\infty a_k^{-s}$, then performing analytic continuation to extend the domain of this function in the complex plane, and then evaluating to compute the value $\zeta_A(-1)$. Alternatively one may define $\zeta_A(s)=\sum_{k=1}^\infty a_kk^{-s}$, perform analytic continuation and then evaluate $\zeta_A(0)$. They are the same for $1+2+3+\cdots$.

It's true $2S=2+4+\cdots$, but this cannot be automatically identified with $0+2+0+4+\cdots$, and that is what must be subtracted from $S=1+2+3+\cdots$ in order to get the desired divergent series $1+0+3+0+5+\cdots$, which itself cannot automatically be identified with the divergent series $1+3+5+\cdots$. These manipulations are not valid for general summability methods.

It does so happen deletion of $0$s is kosher under zeta regularization (the method we're using here), because that's something we can do with series of functions. We're in the realm of holomorphic functions in complex analysis, so let's manipulate some series in that context:

$$\begin{array}{ll} \zeta(s) & = 1^{-s} + 2^{-s}+3^{-s}+\cdots \\ 2^{-s}\zeta(s) & =2^{-s}+4^{-s}+6^{-s}+\cdots \\ (1-2^{-s})\zeta(s) & =1^{-s}+3^{-s}+5^{-s}+\cdots \end{array}$$

However, if we try to subtract the top from the bottom termwise we get

$$-2^{-s}\zeta(s) = 0+(3^{-s}-2^{-s})+(5^{-s}-3^{-s})+(7^{-s}-4^{-s})+\cdots$$

For $s=-1$, the individual terms $3-2$, $5-3$, $7-4$, etc. would all simplify to $1,2,3,\cdots$, however our simplification must be on the level of functions before we can evaluate at $s=-1$, so zeta regularization doesn't provide us with any means of simplifying it to $0+1^{-s}+2^{-s}+\cdots$.