1
$\begingroup$

Given a sequence of complex numbers $(a_n)$ one defines the corresponding Dirichlet series as $$f(s) = \sum_{n = 1}^\infty \frac{a_n}{n^s}.$$ I know that there is some $x_0 \in \mathbb{R} \cup \{\pm \infty\}$ such that $f(s)$ converges absolutely whenever the real part of $s$ is greater than $x_0$.

Suppose that $x_0 < +\infty$ (so $f$ converges absolutely on some half-plane). Is it true that $\lim_{x \to \infty} f(x) = a_1$? If not, does it become true if we know that the sequence $a_n$ does not grow 'too fast' in some sense (for instance exponentially bounded or polynomially bounded)?

Notice that this is a case of switching the order of the limits, because when $x$ goes to infinity, every term in the definition of $f$ goes to zero except the $n = 1$ term. But I'm not sure if switching the limits is allowed in this case.


Motivation: Series like this show up in the theory of finitely generated profinite groups. Specifically, if we calculate the probability that $k$ randomly chosen elements of a fixed finitely generated profinite group $G$ generate this group topologically, we find the answer is given by a Dirichlet series $P_G$ in $k$ with constant term 1. For instance, if $G = \hat {\mathbb{Z}}$ is the group of profinite integers, one finds that $P_G(k)$ is equal to $1/\zeta(k)$, with $\zeta$ the usual Riemann zeta function. If $G$ has a lot of subgroups of finite index, it may happen that $P_G$ fails to converge (i,e, $x_0 = +\infty$ in the notation above). However, if $G$ behaves well enough to guarantee that $P_G$ exists on some half-plane, it would be nice to be able to conclude that $P_G(k)$ goes to $1$ for $k \to \infty$.

$\endgroup$
  • $\begingroup$ Obviously, $f$ might not converge for any $s$. Take $a_n=n^n$ for example. So, first restrict $a_n$. ;-) $\endgroup$ – Mark Viola Feb 11 '16 at 22:02
  • 1
    $\begingroup$ if the Dirichlet series converges somewhere, then yes $\lim_{Re(s) \to +\infty}\sum_n a_n n^{-s} = a_1$ (simply write that if the series converges at $s_0$ then $a_n = \mathcal{O}(n^{s_0})$ and the series converges absolutely for $Re(s) > s_0+1$ ... ) $\endgroup$ – reuns Feb 11 '16 at 22:15
  • $\begingroup$ @user1952009 Thank you! I have worked out the details now, and indeed we get the correct limit. Do you want to post an answer? Otherwise I will probably write an answer myself. $\endgroup$ – Marc Paul Feb 11 '16 at 23:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.