Intuitive question

It is a popular math fact that the sum definition of the Riemann zeta function: $$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} $$ can be extended to the whole complex plane (except one) to obtain $\zeta(-1)=-\frac{1}{12}$. The right hand side for the above equation in $-1$ becomes the sum of the natural numbers so in some sense we have obtained a value for it. My question is: is this value depending on the choice of the Riemann zeta function as the function to be analytically continued, or do we always get $-\frac{1}{12}$?

Proper formulation

let $(f_n:D\subset \mathbb{C}\rightarrow \mathbb{C})_{n\in \mathbb{N}}$ be a sequence of functions and $a \in \mathbb{C}$ with $\forall n\in \mathbb{N}: f_n(a) = n$ and $$f(z):=\sum_{n=0}^\infty f_n(z)$$ convergent on a part of the complex plane, such that it can be analytically continued to a part of the plane that contains $a$. Does it then follow that, under this continuation, $f(a)=-\frac{1}{12}$ and why (or can you give a counterexample)?


  • The case of the Riemann zeta function is the case where $\forall n \in \mathbb{N}: f_n(s) = \frac{1}{n^s}$ and $a=-1$
  • The case where $\forall n \in \mathbb{N}: f_n(z) = \frac{n}{z^n}$ and $a=1$ does yield the sum of all natural numbers but it's continuation $\frac{z}{(z-1)^2}$ has a pole at $a$.
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    $\begingroup$ I have stopped counting how often this "idendity" was presented here. $\endgroup$ – Peter Mar 16 '17 at 23:03
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    $\begingroup$ @Peter Say, it was presented much better than most times, and I can probably count how many times that has happened on my fingers :D $\endgroup$ – Simply Beautiful Art Mar 16 '17 at 23:05
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    $\begingroup$ The comments in this question and the question in general will likely help you much :-) $\endgroup$ – Simply Beautiful Art Mar 16 '17 at 23:08
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    $\begingroup$ Extracted the necessary information from the link above, the answer is no... indeed, if we have $f_n(z,a)=\frac{n}{(n+a)^z}$, then the analytic continuation to $z=0$ yields:$$\lim_{z\to0^+}\sum_{n=1}^\infty f_n(z,a)=\frac{a^2}2-\frac1{12}$$ $\endgroup$ – Simply Beautiful Art Mar 16 '17 at 23:12
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    $\begingroup$ @Peter It happens more than $-\,{1 \over -1/12}$ times. $\endgroup$ – Felix Marin Mar 17 '17 at 0:10

Let $f_n(z,a)=\frac n{(n+a)^z}$. Taking the analytic continuation as $z\to0$ on the sum of $f$, we have

$$\lim_{z\to0^+}\sum_{n=1}^\infty f_n(z,a)=\frac{a^2}2-\frac1{12}$$

The proof is not so bad. Notice that

$$\frac\partial{\partial x}\frac1{(xn+a)^{z-1}}=\frac n{(xn+a)^z}$$

And further that

$$\sum_{n=1}^\infty\frac1{(xn+a)^{z-1}}=\frac1{x^{z-1}}\sum_{n=1}^\infty\frac1{(n+\frac ax)^{z-1}}=\frac{\zeta(z-1,1+\frac ax)}{x^{z-1}}$$

where we use the Hurwitz zeta function. Differentiating with respect to $x$ then gives

$$\frac\partial{\partial x}\frac{\zeta(z-1,1+\frac ax)}{x^{z-1}}=\frac{-ax^{z-3}\zeta(z,1+\frac ax)-(z-1)x^{z-2}\zeta(z-1,1+\frac ax)}{x^{2z-2}}$$

Let $x=1$ and you end up with

$$\sum_{n=1}^\infty f_n(z,a)=-a\zeta(z,1+a)-(z-1)\zeta(z-1,1+a)$$

And as $z\to0^+$, we get...

$$\lim_{z\to0^+}f_n(z,a)=n\\\lim_{z\to0^+}\sum_{n=1}^\infty f_n(z,a)=-a^2\zeta(0)+\zeta(-1)=\frac{a^2}2-\frac1{12}$$


Let $f_n(s) = n^{-s}+ (s+1) e^{-(s+1)n} (b-\zeta(-1))$. Then $f_n(-1) = n$ and for $Re(s) > 1$ : $$F(s) = \sum_{n=1}^\infty f_n(s) = \zeta(s)+(b-\zeta(-1))\frac{s+1}{e^{s+1}-1}$$ It can be continued analytically to the complex plane minus $s=1$ and $s=-1+2ik \pi, k \in \mathbb{Z}^*$ : $$F(-1) = \zeta(-1) + (b-\zeta(-1))\lim_{s \to -1}\frac{s+1}{e^{s+1}-1}=b$$


An analytic continuation of a function is always unique, so long as you have enough information to know the function is analytic. Say you have a function defined for real values of x that are greater than 1. I will use the Riemann Zeta function for our example. That function, zeta(x), is defined for all those x and is analytic for all of those x.

Now, if we find a new function and call it g(x) that equals zeta(x) for all x where zeta is defined, AND, this is the important part. If g(x) is also analytic everywhere, then g(x) is the analytic continuation of zeta(x). It is possible that g(x) won't be defined for all x, but then you could analytically continue g(x).

If you assume that there is a different function h(x), which is also an analytic continuation of zeta(x), then it must be equal to zeta(x) for all x greater than 1. So if take g(x)-h(x) it will be equal to zero, for the entire interval x greater than 1. And any two analytic functions subtracted from one another must give you an analytic function.

The final fact is that if an analytic function is constant on an interval, then the function must be that constant everywhere. So if you have 2 analytic continuations, h(x)-g(x) must equal 0, and h(x)=g(x) must be true.

https://en.wikipedia.org/wiki/Analytic_function#Properties_of_analytic_functions the understanding mainly comes down to learning how analytic functions behave.

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    $\begingroup$ downvoted. Have you read the question? I know what analytic continuation is and why and how it is unique. That was not the question. $\endgroup$ – Jens Renders Jul 26 '18 at 19:00
  • $\begingroup$ Your question was "can we use analytical continuation to obtain sum n=1 to infinity of n such that it doesn't equal -1/12". No you cannot, since the analytic continuation must equal. It doesn't matter what path you take. $\endgroup$ – Sam M Jul 26 '18 at 19:08
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    $\begingroup$ Yes, you did not answer that at all. Note that the sum in question is not a function, the function needs to be chosen and there are many options, as explained in the question en demonstrated in the other answers. $\endgroup$ – Jens Renders Jul 26 '18 at 19:09

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