I was working with some Dirichlet series and I realized that I have never seen any general conditions under which


\begin{equation}
\sum\_{n=1}^{\infty}\frac{a\_n}{n}=\lim\_{s\to1^+}\sum\_{n=1}^{\infty}\frac{a\_n}{n^s}\label{1}\tag{1}
\end{equation}


holds. This is obviously not true in a general case since if so there would be a very simple proof of the PNT from just applying this to $a\_n=\mu(n)$. My question is: under what conditions does \eqref{1} hold?


I can show that if $\sum\_{n=1}^{\infty}\frac{a\_n}{n}$ converges then \eqref{1} must hold using a very simple proof, but I can't find any broader statements. The ideal condition that I would like to show is that if the partial sums $\sum\_{n=1}^{N}\frac{a\_n}{n}$ are bounded then \eqref{1} must hold. I don't know how I would go about proving this though, and any insights on this general area would be greatly appreciated.



Conditions under which $\lim_{s\to1^+}\sum_{n=1}^{\infty}\frac{a_n}{n^s}=\sum_{n=1}^{\infty}\frac{a_n}{n}$
Here are analogous Tauberian theorems for power series and Dirichlet series that involve a condition of *analytic continuation* to a boundary point plus one extra condition that is necessary for the series to converge at some point on that boundary.


Power series: If $c\_n \to 0$ then $\sum\_{n \geq 0} c\_nz^n$ converges for $|z| < 1$. Fatou showed this series converges at each number $z$ with $|z| = 1$ to which the series admits an analytic continuation from inside the disc $\{z : |z| < 1\}$. Note that "$c\_n \to 0"$ is a necessary condition for $\sum c\_nz^n$ to converge at a number $z$ where $|z| = 1$, so including it as a hypothesis is not unreasonable.


Dirichlet series: If $(a\_1 + \cdots + a\_n)/n \to 0$ then $\sum a\_n/n^s$ converges for ${\rm Re}(s) > 1$. Riesz showed this series converges at each $s$ with ${\rm Re}(s) = 1$ to which the series admits an analytic continuation from the half-plane ${\rm Re}(s) > 1$. The condition $(a\_1 + \cdots + a\_n)/n \to 0$ is necessary for $\sum a\_n/n^s$ to converge at a number $s$ where ${\rm Re}(s) = 1$, so including that as a hypothesis is not unreasonable.


For your motivating example with $a\_n = \mu(n)$, of course the hypothesis $(a\_1 + \cdots + a\_n)/n \to 0$ is known by completely elementary methods to be equivalent to the Prime Number Theorem (in the form "$M(x) = o(x)$" for $M(x) = \sum\_{m \leq x} \mu(m)$), so this theorem of Riesz isn't really a good approach to proving the Prime Number Theorem even though the analytic continuation hypothesis for $\sum \mu(n)/n^s = 1/\zeta(s)$ from ${\rm Re}(s) > 1$ to ${\rm Re}(s) = 1$ is valid.


