**Theorem 12** of the following [link](http://empslocal.ex.ac.uk/people/staff/rjchapma/etc/dirichlet.pdf) asserts the following:



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> $\textbf{Theorem.}$ 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, $$\lim\_{s \to 1^{+}} L(s,\chi) = L(1,\chi).$$
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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.



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> * Are there any other nice proofs of the above theorem which you people would like to recommend?
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Doubt in the proof of $\lim_{s \to 1^{+}} L(s,\chi) = L(1,\chi)$
Since you edited the question but did not say that it is clear now, I assume you are hoping for some details in addition to what Ralph said. So:


Let $f\_n(s) = \frac{1}{n^s}- \frac{1}{(n+1)^{s}}$ the derivative of this with respect to $s$ is as you computed $\frac{-\log n}{n^s}+ \frac{\log(n+1)}{(n+1)^{s}}$. This is a function of $s$, and you do not consider it on $[n,n+1]$, but rather $[1,s]$ for each $n$.


The MVT tells you that $\frac{f\_{n}(s) - f\_n(1)}{s -1} = f'(s\_n) $ for some $s\_n\in [1,s]$.
Multiplying by $(s-1)$ and plugging in the explicit expressions for the functions this means
$$(\frac{1}{n^s}- \frac{1}{(n+1)^{s}}) - (\frac{1}{n} -\frac{1}{n+1}) = (s-1)(\frac{-\log n}{n^{s\_n}}+ \frac{\log (n+1)}{(n+1)^{s\_n}}).$$
The left hand side appears in the original sum, and the result is obtained by instead plugging in the right hand side.


