5 Robin Chapman does not actually misuse notation in this ugly way in the article.

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 $$L(s,\chi) = L(1,\chi) + \mathcal{O}(s-1)$$ 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? 4 1+ changed to 1^{+}; edited title Doubt in the proof of \small{\lim_{s\lim_{s \to 1+1^{+}} L(s,\chi) = L(1,\chi)}L(1,\chi) 3 added 42 characters in body; edited title Doubt in the proof of \small\lim_{s\small{\lim_{s \to 1+} L(s,\chi) = L(1,\chi)L(1,\chi)} 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 thatL(s,\chi) = L(1,\chi) + \mathcal{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]$$Biggr] \qquad \qquad \cdots\cdots (1)$$

I don't understand this stephow $(1)$ 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?
2 Inserted the mark (*)
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