# How does $\zeta(1 - s)$ become $(-1/s + \cdots)$?

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

A detailed explanation would be appreciated.

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$\zeta$ 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

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 $s$: $$\zeta(1-s) = -\frac{1}{s} + \gamma + \mathcal{o}(1)$$