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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)?

Examples

  • 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, 2017 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 $\endgroup$ Mar 16, 2017 at 23:05
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    $\begingroup$ The comments in this question and the question in general will likely help you much :-) $\endgroup$ Mar 16, 2017 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:$$\lim_{z\to0^+}\sum_{n=1}^\infty f_n(z,a)=\frac{a^2}2-\frac1{12}$$ $\endgroup$ Mar 16, 2017 at 23:12
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    $\begingroup$ @Peter It happens more than $-\,{1 \over -1/12}$ times. $\endgroup$ Mar 17, 2017 at 0:10

2 Answers 2

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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...

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

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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 $s=-1+2ik \pi, k \in \mathbb{Z}^*$ : $$F(-1) = \zeta(-1) + (b-\zeta(-1))\lim_{s \to -1}\frac{s+1}{e^{s+1}-1}=b$$

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  • $\begingroup$ What could it be that's special about $b = \zeta(-1)$? Some kind of symmetry? $\endgroup$
    – user76284
    Jun 30, 2019 at 3:05
  • $\begingroup$ Actually, there's something fishy about this regularization. The same reasoning applied to a convergent series yields the wrong answer. Let $f_n(s) = n^{-s} + (s - 3)e^{-(s-3)n}(b - \zeta(3))$. Then $f_n(3) = n^{-3}$ and for $\mathrm{Re}(s) > 1$: $F(s) = \sum_{n=1}^\infty f_n(s) = \zeta(s) + (b - \zeta(3)) \frac{s-3}{e^{s-3}-1}$. Then $F(3) = \zeta(3) + (b - \zeta(3)) \lim_{s \rightarrow 3} \frac{s - 3}{e^{s-3} - 1} = b$ which, if $b \neq \zeta(3)$, yields the wrong sum for $\sum_{n=1}^\infty n^{-3}$. $\endgroup$
    – user76284
    Jul 21, 2019 at 9:11
  • $\begingroup$ @user76284 Sure, if you take your regulator with $f_n(3) = n^{-3}$ then $\lim_{s \to 3^+} \sum_n f_n(s) a_n = \sum_n n^{-3} a_n$ whenever $a_n = o(1)$ $\endgroup$
    – reuns
    Jul 21, 2019 at 9:23
  • $\begingroup$ So it seems that, in some sense, $b = \zeta(s)$ is the only legitimate regularization because it leaves fixed the correct values when the sum converges. $\endgroup$
    – user76284
    Jul 21, 2019 at 17:21

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