Question

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?

Answer 1

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.

Answer 2

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.