Abscissa of Convergence (and of Absolute Convergence) of the Derivative of a Dirichlet Series

Question

Given the series:

$$F(s) = \sum f(n) n^{-s}$$

with abscissa of convergence $\sigma_c$. It's derivative would be:

$$F'(s) = - \sum_{n = 1}^\infty \frac{f(n) \log(n)}{n^s}$$

Aopstol, "Intro to Analytic NT" p.236, says the derivative series has the same abscissas of convergence and of absolute convergence as the original series, $F(s)$.

Since the original series is analytic in the half-plane of convergence $\sigma > \sigma_c$, this makes sense.

I was wondering how to show it explicitly for the given series. I have tried assuming every term of the original series is bounded in that half-plane but that only leads to $M \sum_{n = 1}^\infty \log(n)$.

Also I can show by the integral test (just for an initial attempt) that $ \sum_{n = 1}^\infty \frac{\log(n)}{n^s}$ converges.

But in truth, I am stuck. I would appreciate help as to how to show it explicitly and especially that $\sigma > \sigma_c$ is still adequate to take care of the additional $\log (n)$ factor of each term in the derivative series.

Thanks very much.

Answer 1

Apply Apostol's Theorem 11.11 to $|f(n)|$ rather than $f(n)$ to get that the partial sums $$\sum_{n=1}^N |f(n)| n^{-s}$$ converge uniformly on compact subsets lying in the open half-plane to the right of $\sigma_c$. Then apply Lemma 3 (pg. 234) to get that the derivatives $$-\sum_{n=1}^N |f(n)| \log(n) n^{-s}$$ also converge uniformly on compact subsets to the right of $\sigma_c$.

In particular, we get the convergence of $$\sum_{n=1}^N |f(n)| \log(n) n^{-\Re{s}}$$ when $\Re{s} > \sigma_c.$ This is exactly equivalent to the absolute convergence of $$\sum_{n=1}^\infty f(n) \log(n) n^{-s}$$ for $\Re{s} > \sigma_c.$

2nd (and better!) method requested by Andrew

Let $\sigma$ be greater than $\sigma_c$. Choose $\sigma_2$ such that $\sigma_c < \sigma_2 < \sigma$. Then $\sigma_2$ is in the region of absolute convergence of $F$, we have $$\sum_{n=1}^\infty \frac{|f(n)|}{n^{\sigma_2}} < \infty.$$ Now let $\epsilon > 0$. No matter how small $\epsilon$ is, for sufficiently large $n$ we have that $n^\epsilon > \log(n)$. Thus,

$$\sum_{n=1}^\infty \frac{|f(n)|}{n^{\sigma_2}} \frac{\log(n)}{n^\epsilon} < \infty.$$ In particular, for $\epsilon = \sigma - \sigma_2$, we have $$\sum_{n=1}^\infty \frac{|f(n)|\log(n)}{n^{\sigma}} < \infty.$$ Thus, $$\sum_{n=1}^\infty \frac{f(n)\log(n)}{n^{s}}$$ converges absolutely for any $s$ with $\Re{s} > \sigma_c$.

Answer 2

Up to Encyclopedia of Mathematics, the abscissa of the absolute convergence of a Diriclet series $$A=\limsup_{n\to\infty} \frac {\log |a_n|} {\lambda_n}.$$ In your case for $F'(s)$ $a_n= - f(n)\log(n) $ and $\lambda_n=- \log (n).$ Therefore, $$ A= \lim_{n\to \infty}\frac {\log|f(n)\log(n)|} {-\log (n)}=\lim_{n\to \infty}\frac {\log|f(n)|} { -\log (n)}+\lim_{n\to \infty}\frac {\log|\log(n)|} {- \log (n)}=\sigma_c+0=\sigma_c$$ in your notation.