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

Solution 1:

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 $z=1$.

Solution 2:

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

Solution 3:

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.

Solution 4:

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.