# Showing that $\displaystyle\lim_{s \to{1+}}{(s-1)\zeta(s)}=1$

I need prove the following: ($\zeta(s)$ is the Riemann zeta function)

$\displaystyle\lim_{s \to{1+}}{(s-1)\zeta(s)}=1$

I really don't know, i have tried, but nothing for now.

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Hint: $$\zeta(z)(1-2^{1-z})=1^{-z}-2^{-z}+3^{-z}-4^{-z}+\dots$$ Therefore, $$\lim_{z\to1}(z-1)\zeta(z)=\lim_{z\to1}\frac{z-1}{1-2^{1-z}}\ \lim_{z\to1}\,(1^{-z}-2^{-z}+3^{-z}-4^{-z}+\dots)$$ Note added: $1^{-z}-2^{-z}+3^{-z}-4^{-z}+\dots$ converges for $\mathrm{Re}(z)\gt0$, thus providing an analytic continuation of $\zeta$ to the entire right half-plane except $z=1$.

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Hint: Approximate the series for $\zeta(s)$ above and below by integrals.

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For $s>1$, we have \begin{align} \zeta(s) & = \int_{1^-}^{\infty} \dfrac{d \lfloor x \rfloor}{x^s} = \left.\dfrac{\lfloor x \rfloor}{x^s} \right \vert_{x=1^-}^{\infty} +s \int_{1^-}^{\infty} \dfrac{\lfloor x \rfloor}{x^{s+1}} dx = s \int_{1^-}^{\infty} \dfrac{\lfloor x \rfloor}{x^{s+1}} dx\\ & = s \left(\int_{1^-}^{\infty} \dfrac{dx}{x^{s}} - \int_{1^-}^{\infty} \dfrac{\{x\}dx}{x^{s+1}} \right) = \dfrac{s}{s-1} - s\int_{1^-}^{\infty} \dfrac{\{x\}dx}{x^{s+1}} \end{align} Hence, we get that $$(s-1)\zeta(s) = s - s(s-1)\int_{1^-}^{\infty} \dfrac{\{x\}dx}{x^{s+1}}$$ Now argue why $\displaystyle \lim_{s \to 1} s(s-1)\int_{1^-}^{\infty} \dfrac{\{x\}dx}{x^{s+1}} =0$ and finish it off.

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I've never studied the Zeta function in depth, and so I've never seen it represented this way - do you have any writeups on this representation handy? Thanks – DanZimm Jun 6 '13 at 3:37
The representation isn't so weird: think about what it means to integrate against $d\lfloor x \rfloor$. – A Walker Jun 6 '13 at 3:43
@DanZimm The answer by Eric and the comment by Zev has the desired details. – user17762 Jun 6 '13 at 3:46
@user17762 OH duh! forgot that the floor operator can be thought of as a function xD – DanZimm Jun 6 '13 at 3:54
@RandomVariable Yes. And this can be extended to $\text{Re}(s)>-n$ where $n \in \mathbb{Z}^+$, by performing integration by parts $n$ times. – user17762 Jun 6 '13 at 4:33

Skectched proof (I think something like this may be what Antonio meant):

$$\text{For}\;\;x>0\,,\,s>1\;,\;\; f(x):=\frac1{x^s}\;\;\text{is monotone descending}$$

$$\text{For}\;\;n\in\Bbb N\;,\;\;\min_{x\in[n,n+1]}\frac1{x^s}\le\int\limits_n^{n+1}\frac{dx}{x^s}\le\max_{x\in[n,n+1]}\frac1{x^s}\implies$$

$$\frac1{(n+1)^s}\le\frac1{1-s}\left(\frac1{(n+1)^{1-s}}-\frac1{n^{1-s}}\right)\le\frac1{n^s}$$

Sum over $\,n\in\Bbb N\,$ (Note the telescopic series!):

$$\sum_{n=1}^\infty\frac1{(n+1)^s}\le\frac1{1-s}\sum_{n=1}^\infty\left(\frac1{(n+1)^{1-s}}-\frac1{n^{1-s}}\right)\le\sum_{n=1}^\infty\frac1{n^s}\implies$$

$$\zeta(s)-1\le\frac1{s-1}\le\zeta(s)\implies1\le(s-1)\zeta(s)\le s$$

and squeeze theorem and we're done.

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+1, precisely what I was hinting at. – Antonio Vargas Jun 7 '13 at 12:35