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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
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(1) Your link shows nothing about $\,\zeta\,$ , though I guess this must be Riemann's zeta function, and (2) Wow, 0% accept rate? This would make Riemann's conjecture go contradictory. Either you don't accept answers given to your past questions because you don't like them (and then what to ask here for?), or else you...just haven't done that. I very much advice you to accept answers as this will increase your qeustion's chances to be addressed. –  DonAntonio 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
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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
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Thanks for increasing your accept rate. It has little to do with reputation. Please see the various discussions of accepting answers on the meta site. –  Gerry Myerson Sep 30 '12 at 10:27

1 Answer 1

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

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Great answer! Thanks. –  glebovg Sep 30 '12 at 8:07

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