**Theorem 12** of the following link asserts the following:

$\textbf{Theorem.}$ Let $\chi \in X_{N}$ with $\chi \neq \epsilon$. There exists $C > 0$ such that $$L(s,\chi) = L(1,\chi) + O(s-1)$$ as $s \to 1^{+}$. In particular, $$\lim_{s \to 1^{+}} L(s,\chi) = L(1,\chi).$$

The proof is as follows: Let $1< s < 2$. From the proof of $\textbf{Theorem 9}$ we have $$L(s,\chi) - L(1,\chi) = \sum\limits_{n=1}^{\infty} a_{n} \Biggl[\biggl(\frac{1}{n^s} - \frac{1}{(n+1)^{s}}\biggr) - \biggl(\frac{1}{n} - \frac{1}{n+1}\biggr)\Biggr]$$ where the sequence $\{a_{n}\}$ is bounded. Applying the mean value theorem to the function $s \mapsto n^{-s} - (n+1)^{-s}$ gives a sequence $\{s_{n}\}$ with $1 < s_{n} < s$ and $$L(s,\chi) - L(1,\chi) = (s-1) \sum\limits_{n=1}^{\infty} a_{n} \Biggl[\frac{\log\:(n+1)}{(n+1)^{s_n}} - \frac{\log\:(n)}{n^{s_n}} \Biggr] \qquad \qquad \cdots\cdots (1)$$

I don't understand how $(1)$ is derived. When I applied the Mean-Value-Theorem to the function $f(s)=x^{-s} - (x+1)^{-s}$ on $[n,n+1]$ i get $$f'(s) = -x^{-s}\log\:(x) + (x+1)^{-s}\log\:(x+1).\hspace{40pt}(\ast)$$ So by the Mean-Value-Theorem i get an $s_{n} \in (n,n+1)$ such that $$f'(s_{n}) = -n^{-s_n}\log\:(n) + (n+1)^{-s}\log\:(n+1) - (n+1)^{-s_n}\log\:(n+1) + (n+2)^{-s_n}\log\:(n+2)$$ which gives \begin{align*} f'(s_{n}) &= \frac{\log(n+2)}{(n+2)^{s_n}} - \frac{\log\:(n)}{n^{s_n}} \\\ &= \frac{f(b)-f(a)}{b-a} = (n+1)^{-s} - (n+2)^{-s} - (n+1)^{-s} + n^{-s} \\\ &= \frac{1}{n^s} - \frac{1}{(n+2)^s} \end{align*}

Am I making a mistake. I am not able to see how the author get's to that step.

- Are there any other nice proofs of the above theorem which you people would like to recommend?