Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Why is $$\zeta(1 - s) = -\frac{1}{s} + \cdots$$ for small negative values of $s$?

A detailed explanation would be appreciated.

share|cite|improve this question
$\zeta$ is meromorphic with a simple pole at $s=1$. – anon Sep 30 '12 at 4:14
@DonAntonio $\zeta(1-s)$ gets turned into a series expansion between the last two lines in the linked article. – anon Sep 30 '12 at 4:15
Thanks @anon , the link works now. – DonAntonio Sep 30 '12 at 4:19
@anon How? Can you show? – glebovg Sep 30 '12 at 4:21
A meromorphic function $f(z)$ with a simple pole at $z=a$ with residue $r$ admits (locally) a power series expansion $$f(z)=\frac{r}{z-a}+c_0+c_1(z-a)+\cdots.$$ For reference, consult, say, any text or comprehensive notes on complex analysis / variables / functions. The fact that zeta is meromorphic with simple pole of residue $1$ at $s=1$ is often considered common knowledge (it should be listed on just about any serious reference on the Riemann zeta function on the web), and the proof contained in any good introductory notes on $\zeta$ - or, for instance, the link I gave. – anon Sep 30 '12 at 4:44
up vote 8 down vote accepted

For $s < 0$ $$ \begin{eqnarray} \zeta(1-s) &=& \sum_{n=1}^\infty n^{s-1} = \int_1^\infty x^{s-1} \mathrm{d}x + \sum_{n=1}^\infty \left( n^{s-1} - \int_{n}^{n+1} x^{s-1} \mathrm{d}x \right) \\ &=& -\frac{1}{s} + \sum_{n=1}^\infty \left( n^{s-1} - \frac{(n+1)^s - n^s}{s} \right) \end{eqnarray} $$ Now, observe that the following limit is finite $$ \lim_{s \to 0^-} \zeta(1-s) + \frac{1}{s} = \sum_{n=1}^\infty \left( \frac{1}{n} - \log\frac{n+1}{n} \right) = \gamma \tag{1} $$ where $\gamma$ is the Euler-Mascheroni constant. The sum on the right-hand-side of $(1)$ converges , since for large $n$ $$ \frac{1}{n} - \log\left( 1+\frac{1}{n}\right) = \frac{1}{2 n^2} + \mathcal{o}\left(\frac{1}{n^2}\right) $$ Alternatively: $$ \sum_{n=1}^\infty \left( \frac{1}{n} - \log\frac{n+1}{n} \right) = \lim_{m \to \infty} \sum_{n=1}^m\left( \frac{1}{n} - \log\frac{n+1}{n} \right) = \lim_{m \to \infty} \left( \sum_{n=1}^m \frac{1}{n} - \log(m+1) \right) = \gamma $$ Hence, this establishes that, for small negative $s$: $$ \zeta(1-s) = -\frac{1}{s} + \gamma + \mathcal{o}(1) $$

share|cite|improve this answer
Great answer! Thanks. – glebovg Sep 30 '12 at 8:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.