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Suppose we have a Dirichlet series

$$ D(s) = \sum_{n=1}^\infty \frac{a(n)}{n^s} $$

which we know is absolutely convergent for $Re(s)>1$. Suppose that we prove that $\lim_{s\to 1^+}D(s) < \infty$. Does this imply that $D(s)$ continues analytically to some half plane to the left of $s=1$, that is to the region $Re(s) > 1-\epsilon$ for some positive $\epsilon$?

Remark: The case where $a(n)>0$ for all $n$ (or equivalently finitely many $n$) is a classical theorem of Landau. I am interested when this is not the case.

Krishnarjun
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  • Remark: Landau's theorem is more subtle than what you say -- it is not enough that $D(1+)=\infty$ as you write. Here is one counterexample https://math.stackexchange.com/questions/2322161/landaus-theorem-dirichlet-series. – D.R. Feb 20 '23 at 06:26

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This is just a comment that seems important....

Note that $$\frac{\zeta(2s)}{\zeta(s)}=\sum_{n=1}^{\infty}\frac{\lambda(n)}{n^s},$$ for Re$(s)>1$ and that $\frac{\zeta(2s)}{\zeta(s)}=\sum_{n=1}^{\infty}\frac{\lambda(n)}{n^s}=0$ at $s=1$ (this is equivalent to the prime number theorem). So, your question includes the question of whether $\frac{\zeta(2s)}{\zeta(s)}$ can be analytic continued to a half-plane Re$(s)>1-\epsilon$, That is, Is there $\epsilon>0$ such that $\zeta(s)\neq 0$ for Re$(s)>1-\epsilon$? This is an open problem, about the zeros of $\zeta(s)$ in the critical strip $0<$Re$(s)<1$.

GoldSoundz
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  • Very nice observation. It seems that the question is much harder than what I had originally anticipated. – Krishnarjun Jun 01 '22 at 09:18
  • The exact same argument can be applied to $1/\zeta(s) = \sum_{n\geq 1} \mu(n)/n^s$, which is arguably more basic as it has constant numerator $1$ instead of $\zeta(2s)$. – KCd Feb 22 '24 at 22:34