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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?
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If I'm not missing something, it looks fine. For, let $f(s) = n^{-s} - (n+1)^{-s}$. Then $f(s) - f(1) = (s-1)f'(s_n)$. Setting in your computation of $f'$ in $(\ast)$ gives just the formula over the phrase "I don't understand ...". –  Ralph Feb 19 '12 at 14:52
Also note that the paper wants $s_n \in (1,s)$ and not $s_n \in (n,n+1)$ (see the line after the $(\ast)$). –  Ralph Feb 19 '12 at 15:24
@Ralph: Dear Ralph, thanks for your response. I shall work through this and get back to you. –  Chandrasekhar Feb 19 '12 at 16:12
Why the downvote without a reason. –  Chandrasekhar Feb 20 '12 at 10:38
The downvote was probably because your question was not of research level. You asked for clarifying a certain step in a certain proof of a well-known fact. E.g. see my response below for two more proofs available in introductory textbooks. –  GH from MO Feb 20 '12 at 19:44
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2 Answers

up vote 3 down vote accepted

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.

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@quid: Thanks a lot for the help. –  Chandrasekhar Feb 20 '12 at 10:39
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It is a standard fact that $L(s,\chi)$, initially defined as a locally uniformly convergent Dirichlet series in $\Re s>1$, extends to a holomorphic function on $\mathbb{C}$. See for example Chapter 9 in Davenport: Multiplicative Number Theory, especially the first half of Page 69.

Now basic complex analysis tells us that the Taylor series of $L(s,\chi)$ around $s=1$ converges absolutely on $\mathbb{C}$, of which $$ L(s,\chi)=L(1,\chi)+O(|s-1|) $$ is a consequence. In fact for the last equation we only need to know that $L'(s,\chi)$ is continuous on $\mathbb{C}$, which follows directly from Cauchy's integral formula.

A quick proof of the holomorphicity of $L(s,\chi)$ in $\Re s>0$ follows from the fact that the partial sums $\sum_{n\leq x}\chi(n)$ are bounded. See Proposition 9 in Section VI.2 of Serre: A course in arithmetic, or Theorem 1.3 in Montgomery-Vaughan: Multiplicative number theory I.

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@GH: Thanks for your answer. –  Chandrasekhar Feb 20 '12 at 20:21
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