The Riemann Zeta Function is defined as $ \displaystyle \zeta(s) = \sum\limits_{n=1}^{\infty} \frac{1}{n^s}$. It is not absolutely convergent or conditionally convergent for $\text{Re}(s) \leq 1$. Using analytic continuation, one can derive the fact that $\displaystyle \zeta(-s) = -\frac{B_{s+1}}{s+1}$ where $B_{s+1}$ are the Bernoulli numbers. Can one obtain this result without resorting to analytic continuation?

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    $\begingroup$ What is $B_s$?. $\endgroup$ Nov 3, 2010 at 3:27
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    $\begingroup$ I suppose Trevor means the Bernoulli numbers. $\endgroup$
    – user02138
    Nov 3, 2010 at 3:28
  • $\begingroup$ So you don't want to use the reflection relation? $\endgroup$ Nov 3, 2010 at 4:04
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    $\begingroup$ If you do not do analytic continuation, what do you mean by $\zeta(s)$ for negative integer $s$? $\endgroup$ Nov 3, 2010 at 4:13
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    $\begingroup$ Your formula for zeta at -s should not use the variable s. There you specifically mean s is a positive integer. Try zeta(-k) = -B_{k + 1}/(k + 1) where k is a positive integer. To answer your main question, you simply need some way to construct an analytic formula for the zeta-function that makes sense beyond the half-plane where the series is defined. There may be ways to do this without explicitly using the magic words "analytic continuation", but unless you're Euler any method you use is likely to provide an analytic continuation. $\endgroup$
    – KCd
    Nov 3, 2010 at 9:45

2 Answers 2


Using Euler--MacLaurin summation, one can obtain the following formula for $\zeta(s)$:

$$ \zeta(s) = \frac{1}{s-1}+\frac{1}{2} + \frac{B_2}{2} s + \cdots \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $$ $$ \cdots + \frac{B_{2k}}{(2k)!}s(s+1)\cdots (s + 2k-2) + \frac{s(s+1)\cdots(s+2k-1)}{(2k)!}f(s), $$ where $f(s)$ is an integral involving $s$ which converges when $\Re(s) > -2k$. (My favourite reference for $\zeta(s)$ is Edwards's book Riemann's zeta function. This particular formula is gotten by setting $N = 1$ in formula (1) on p. 114.)

So this gives a formula for $\zeta(s)$ which is defined when $\Re(s) > -2k$. If you substitute in $s = -2k+1$, you will get (after some rearrangement) that $\zeta(-(2k-1)) = -B_{2k}/2k.$

Of course this is a form of anaytic continuation (as others have noted, it is hard to make sesne of what $\zeta(-(2k-1))$ would mean otherwise). But it perhaps a little different to the standard approach.

ADDED: An approach which seems quite different to analytic continuation --- at least at first --- is the Abelian regularization approach used by Euler. (Please excuse the anachronism of labelling Euler's method with Abel's name!) This is disussed in some of the answers to this question.

The idea is first to multiply by $(1-2^{-s+1})$, which eliminates the pole at $s=1$, and replaces $\zeta(s)$ by the function $\eta(s):= \sum_{n=1}^{\infty} (-1)^{n-1} n^{-s}$. (Clearly if we can evaluate $\eta(s)$, we can evaluate $\zeta(s)$, simply by divising through by $(1-2^{-s+1})$.) Then, one computes $\eta(-k)$ via the following formula: $$\eta(-k) = \lim_{T \to 1} \sum_{n=1}^{\infty}(-1)^{n-1}n^k T^n.$$ The point is that the series in $T$ converges (when $|T| < 1$) to a rational function of $T$, which we can then evaluate at $T = 1$.

This method can be seen directly to lead to the usual formula in terms of Bernoulli numbers. One can also relate it to the usual description of $\zeta(s)$ (or --- equivalently --- $\eta(s)$) via analytic continuation (by considering the two variable function $\sum_n (-1)^n n^{-s} T^n$), but it is the approach I know which is a priori furthest removed from analytic continuation.

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    $\begingroup$ I felt the need to express my strong support here : Edward's book is indeed very useful, and cheap! $\endgroup$
    – Joel Cohen
    Jun 28, 2011 at 14:46
  • $\begingroup$ I'm confused about the use of Euler--MacLaurin summation formula for $N=1$, in this case we have $\zeta(s)=\sum_{n=1}^{N-1}n^{-s}+\cdots$ or $\zeta(s)=\sum_{n=1}^{0}n^{-s}+\cdots$ wich leads to division by $0$!!! I'sure I'm missing something but I don't know what... could you explain? $\endgroup$
    – Neves
    May 6, 2013 at 17:46
  • $\begingroup$ @Neves: Dear Neves, The case $N=1$, doesn't lead to division by zero. Rather, $\sum_{n = 1}^0$ is simply the empty sum, whose value equals $0$. Regards, $\endgroup$
    – Matt E
    May 6, 2013 at 17:50
  • $\begingroup$ Could we say that $\zeta(s)=\frac{1}{s-1}+\frac{1}{2}+\sum_{k=1}^{\infty}\frac{B_{2k}}{(2k)!}x^{(2k)}$ is a válid representation for $\zeta(s)$ in the wole complex plane? $\endgroup$
    – Neves
    May 7, 2013 at 17:24
  • $\begingroup$ @Neves: Dear Neves, No, I don't think so. This is an asymptotic expansion, not a convergent series (as far as I remember). See the discussion in Edwards's book for more details. Regards, $\endgroup$
    – Matt E
    May 7, 2013 at 17:44

What are complex Bernoulli numbers?


Bernoulli numbers are only defined for natural numbers.

So you probably mean $s$ being an integer. As KCrad points out in the comments: Euler has proven a functional equation for the Riemann zeta function before Riemann using integrals and only for real numbers, so that is a method.

But if you plugin complex values in the integral, you are back to analytic continuation, so the answer is most likely no.

E.g., do you have a nice heuristic why $\zeta(-1) =1+2+3+\dots = 1/12$?

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    $\begingroup$ Bit late, but for future readers: $\zeta(-1)\not = 1+2+3+\dots$, this exactly why we need to analytically continue past the halfplane $Re(z)>1$. $\endgroup$ Sep 27, 2018 at 19:48

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