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 hard, but have obtained nothing for now.

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.
