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Apologies if this has been asked already. Inspired partly by this answer where an $n e^{-\epsilon n}$ rather than $n^s$ regularization was made in the 'evaluation' of $\sum\limits_{n=1}^{\infty}n$ and the number $-\frac{1}{12}$ appeared as the only constant in the answer, and partly by a conversation with a friend claiming that $-\frac{1}{12}$ is the only 'right' way of summing this series, I have the following question:

Does anyone know of or can think of a function $p(s,n)$, $s\in\mathbb{C}$, $n\in\mathbb{N}$ such that $\exists s_0\in\mathbb{C}$ such that $p(s_0,n)=n$, $\forall n\in\mathbb{N}$, where $q(s)=\sum\limits_{n=1}^{\infty}p(s,n)$ is defined and analytic on some domain $D\in\mathbb{C}$ (not containing $s_0$ obviously) and such that $q(s)$ can be analytically continued to $Q(s)$ defined on some larger domain $D_0\supset D$ such that $s_0\in D_0$ and $Q(s_0)\ne-\frac{1}{12}$?

I expect that finding such a function is probably simple, but I have not been successful so far, partly due to not having much familiarity with analytic continuations. (Note that I am not wishing to make any claims or have any arguments about the validity of $1+2+3+4+...\stackrel{?}{=}-\frac{1}{12}$).

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Anon
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  • @user1952009 I see, that was easier than I expected. All we have to do is take a separate function and add it as long as it also converges in the desired region, and goes to zero at our point, we just have to prove analyticity. With your function we get $Q(s)=\zeta(s)+(s+1)\zeta(s+1)$. – Anon Nov 01 '16 at 05:59
  • I wonder if there's anything special in particular about the functions $n^s$ and $\zeta(s)$. – Anon Nov 01 '16 at 06:10
  • Wait, but won't this still give $Q(-1)=-\zeta(-1)+0\zeta(0)=-\frac{1}{12}$? So this doesn't seem to be what I was asking for after all? – Anon Nov 01 '16 at 06:20
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    Yes sorry use instead $p(s,n) = n^{-s} + (s+1) n^{-s-2}$ so that $Q(-1) = \lim_{s \to -1} \zeta(s)+(s+1) \zeta(s+2) = \frac{-1}{12}+\lim_{s \to 1} (s-1)\zeta(s) = \frac{11}{12}$ – reuns Nov 01 '16 at 06:23

2 Answers2

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$-1/12$ is not really special among the analytic continuation methods, since $$p(s,n) = n^{-s}+(s+1)n^{-2-s} \implies Q(s) = \sum_{n=1}^\infty p(s,n) = \zeta(s)+(s+1) \zeta(s+2) \qquad(Re(s) > 1) $$ $$\implies Q(-1) = \lim_{s \to -1} Q(s)= \zeta(-1)+\lim_{s \to 1}(s-1)\zeta(s)= \frac{11}{12}$$

Thus a natural question is what is special with the $\zeta$ regularization ?

  • The $z^n$ regularization is easier to understand : If $$f(z) = \sum_{n=-\infty}^\infty a_n z^n$$ converges for $r < |z| < R$ and can be analytically continued to $z=1$, then $\lim_{z \to 1} f(z) = \lim_{z \to 1} z^k f(z)$ that is $(a_n)$ is the same sequence as $(a_{n-k})$ with respect to power series regularization.

  • Now yes the $n^{-s}$ regularization is special in how it acts on shifting the sequence (in a complicated way). But with $$g(s) = \sum_{n=1}^\infty a_n n^{-s}$$ you get that $\lim_{s \to 0} g(s) = \lim_{s \to 0}g(sk)=\lim_{s \to 0} m^{-s}g(s)$ that is $(a_n)$ is the same sequence as $(a_{n^{1/k}})$ and $(a_{n/m})$ for the zeta regularization, and sometimes it is more useful than shifting invariance (multiplicative functions..).

reuns
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    I can't upvote your answer right now 'cos I've reached the daily limit. I've thus accepted it since it nicely answers the basics of my question (although it would probably be just as well to mention why $Q$ is analytic and can be continued in the same way since it is a sum of two $\zeta$ functions). – Anon Nov 01 '16 at 06:37
  • How do you get $\lim_{s \to 1}(s-1)\zeta(s)=1$ ? – Takahiro Waki Nov 01 '16 at 08:28
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    @TakahiroWaki see https://en.wikipedia.org/wiki/Riemann_zeta_function#Definition – reuns Nov 01 '16 at 08:31
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    @Anon I feel you all the time. Too many good things for my limited voting rights. – Simply Beautiful Art Dec 10 '16 at 13:53
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Here's a fun one:

$$\zeta^\star(s,\alpha)=\sum_{n=1}^\infty\frac n{(n+\alpha)^s}$$

Analytically continuing to $s=0$, you'll get

$$\zeta^\star(0,\alpha)=\frac{\alpha^2}2-\frac1{12}$$

And likewise,

$$n=\frac n{(n+\alpha)^0}\forall\alpha$$

This result may be derived by considering another variable:

$$\sum_{n=1}^\infty\frac1{(nx+\alpha)^s}$$

And differentiating w.r.t. $x$. Then factor $x$ out and see the resulting Hurwitz zeta function, which may easily be handled, then letting $x=1$.

Simply Beautiful Art
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  • Just to complement this answer, I worked the details and I got $\sum_{n\ge1}\frac{n}{(n+\alpha)^s}=\zeta(s-1,\alpha)-\alpha\zeta(s,\alpha)$. Strictly valid for $\Re(s)>2$ I think. – Enredanrestos Feb 16 '18 at 00:08