TITLE: Showing that $\displaystyle\lim_{s \to{1+}}{(s-1)\zeta(s)}=1$
QUESTION [12 upvotes]: 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 hard, but have obtained nothing for now.

REPLY [7 votes]: Sketched 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.