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I saw NumberPhile channel on Youtube, and they proved $1+2+3+\cdots=-1/12$. Also, I read This.

So, which one is correct

$$\zeta(-1)=-1/12\\ \text{or} \\\zeta(-1) \to -1/12$$

Equivalent to:

$$1+2+3+\cdots=-1/12\\ \text{or} \\1+2+3+\cdots \to -1/12$$

My question: Does it "equal" or "converge"?

Question Explanation:

I mean by "$\to$" "approaches to", like $x\to a $ means $\forall \epsilon>0, |x-a|<\epsilon.$

marked as duplicate by nbubis, Grigory M, Davide Giraudo, Peter Košinár, Dan Rust Jan 9 '14 at 18:45

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  • 2
    Related and recent:… – Henry Jan 9 '14 at 17:06
  • Explanation:"=" is a particular case of "$\to$". For example $x=a \implies x\to a$ but the converse is not always true. – AHH Jan 9 '14 at 17:38
  • @Henry , there is a difference between this question and "related question". My question: Does it "equal" or "converge"?. – AHH Jan 9 '14 at 18:37

For the zeta function, it is correct that


The function $\zeta$ is continuous (and continuous in $-1$ in particular).

However, with the usual notion of convergence, we have:

$$1+2+3+4+\cdots = +\infty$$

which is the same as

$$1+2+3+4+\cdots+n \to +\infty \quad\mathrm{as}\quad n\to+\infty$$

To say that

$$1+2+3+4+\cdots = -\frac{1}{12}$$


$$1+2+3+4+\cdots+n \to -\frac{1}{12} \quad\mathrm{as}\quad n\to+\infty$$

would require us to specify in what sense this limit is taken. See Wikipedia on $1+2+3+4+\cdots$ for details.

  • "=" is a particular case of "$\to$". For example $x=a \implies x\to a$ but the converse not always true. – AHH Jan 9 '14 at 17:32

$\zeta(z)$ is a meromorphic function with a single pole, which has residue $1$ at $z=1$. $\zeta(-1)=-\frac1{12}$. This fact is used to justify the divergent series $$ 1+2+3+4+\dots=-\frac1{12} $$ but that divergent series is not why we say that $\zeta(-1)=-\frac1{12}$.

Computation of $\mathbf{\zeta(-1)}$:

Multiply equation $(1)$ from this answer by $x+1$, then integrate by parts twice, to get $$ \begin{align} (1-2^{1-x})\zeta(x)\Gamma(x+2) &=\int_0^\infty\frac{(x+1)xt^{x-1}}{e^t+1}\mathrm{d}t\\ &=\int_0^\infty\frac{(x+1)t^xe^t}{(e^t+1)^2}\mathrm{d}t\\ &=\int_0^\infty\frac{t^{x+1}(e^{2t}-e^t)}{(e^t+1)^3}\mathrm{d}t\tag{1} \end{align} $$ Now we can plug in $x=-1$ into $(1)$ to get $$ \begin{align} (1-2^2)\zeta(-1)\Gamma(1) &=\int_0^\infty\frac{e^{2t}-e^t}{(e^t+1)^3}\mathrm{d}t\\ &=\int_1^\infty\frac{u-1}{(u+1)^3}\mathrm{d}u\\ &=\int_1^\infty\left(\frac1{(u+1)^2}-\frac2{(u+1)^3}\right)\mathrm{d}u\\ &=\frac14\tag{2} \end{align} $$ Since $(1-2^2)\Gamma(1)=-3$, $(2)$ says that $$ \zeta(-1)=-\frac1{12}\tag{3} $$

It could make sense to say that $\zeta(z)\to-1/12$ as $z\to-1$, but the statement $\zeta(-1)\to-1/12$ as $\text{what}\to\text{what?}$ can make sense only if $\zeta(z)$ is actually equal to $-1/12$.

As for the statement that $1+2+3+\cdots$ bears some relation to $-1/12$, that requires one to define one's "summation method". But even there, one cannot say $1+2+3+\cdots\to\text{something}$ without at least tacitly saying "$\text{as something}\to\text{something}$, and if the variable that's approaching something does not appear in the expression $1+2+3+\cdots$, then the limit statement would be true only if $1+2+3+\cdots$ is actually equal to the supposed limit.

Later postscript: In view of comments below, may I ask that those who want to comment read what I ACTUALLY wrote before commenting on it?

  • 1
    This isn't right. A (discontinuous) function can have a well-defined limit approaching a point and then take a different value at that point. In this case, $\zeta(-1)=-1/12$, but $\zeta(z)=\sum_{n}n^{-z}$ only when $\text{Re}(z)>1$. – mjqxxxx Jan 9 '14 at 17:05
  • 1
    @mjqxxxx : I think you need to read carefully what I wrote. Suppose $f$ is a discontinuous function and $f(5)=8$. Then it may be that $f(x)\to\text{something}\ne8$ as $x\to5$, but it makes no sense to say $f(5)\to\text{something}$ unless $f(5)$ is actually equal to that number. – Michael Hardy Jan 9 '14 at 17:07
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    @mjqxxxx : Really, your reading comprehension is very very poor in this instance. – Michael Hardy Jan 9 '14 at 17:08
  • That's fine. Your point is just that OP's notation (and understanding of what $\rightarrow$ means) is flawed. $\zeta(-1)\rightarrow -1/12$ (in any limit you like, since there's no variable in sight) is equivalent to $\zeta(-1)=-1/12$. I was just assuming that his intended question was whether $\zeta(z)\rightarrow -1/12$ as $z\rightarrow -1$ or $\zeta(-1)=-1/12$ (or both); but you may be right that his confusion is more fundamental. – mjqxxxx Jan 9 '14 at 18:29

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