TITLE: Conditions under which $\lim_{s\to1^+}\sum_{n=1}^{\infty}\frac{a_n}{n^s}=\sum_{n=1}^{\infty}\frac{a_n}{n}$
QUESTION [9 upvotes]: 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.

REPLY [2 votes]: In D. J. Newman's paper 
A simple analytic proof of the prime number theorem 
published in The American Mathematical Monthly
87 (1980) 693-696, Newman proved a result of this type which may also be useful to you (this is a minor variant of the criterion given by KConrad in his answer).
It says the following:
Let $|a_n| \leq 1$ and suppose that the Dirichlet series 
$$\sum_{n = 1}^\infty \frac{a_n}{n^s}$$
admits a holomorphic continuation to the line $\mathrm{re}\, s = 1$. Then 
$$\sum_{n = 1}^\infty \frac{a_n}{n^s}$$
converges for all $\mathrm{re}\, s \geq 1$.