Let $\alpha(s) = \sum_{n = 1}^{\infty} a_n n^{-s}$ be a Dirichlet series with abscissa of convergence as $\sigma_c$. Then it is natural to think that $\alpha(s)$ must have some kind of singularity on the line $\sigma = \sigma_c$ which causes the Dirichlet series $\alpha(s)$ to diverge to the left of $\sigma_c$. It turns out that this is partially correct. Landau showed if the coefficients $a_n$ are nonnegative (for sufficiently large $n$), then $\alpha(s)$ has a singularity at the point $s = \sigma_c$. However, if not all $a_n$ are nonnegative, then in some cases $\alpha(s)$ can in fact be analytically continued to (an open set containing) $\sigma_c$. For instance, if $q > 1$ and $\chi$ is not a principal character modulo $q$, then the Dirichlet $L$-function
\[
L(s, \chi) = \sum_{n = 1}^{\infty} \chi(n) n^{-s}
\] can be analytically continued to the entire complex plane. We now state and prove Landau’s amazing theorem.
Theorem. (Landau) Let $\alpha(s) = \sum_{n = 1}^{\infty} a_n n^{-s}$ be a Dirichlet series with finite abscissa of convergence $\sigma_c$. If $a_n \geq 0$ for every $n$, then $\sigma_c$ is a point of singularity of $\alpha(s)$.
We remark here that it suffices to assume that $a_n \geq 0$ for all sufficiently large $n$ since $\sum_{n \leq N} a_n n^{-s}$ is an entire function for a fixed $N$.
Proof. Suppose for the sake of contradiction that $\sigma_c$ is not a point of singularity of $\alpha(s)$, i.e., there exists an analytic continuation of $\alpha(s)$ to an open set $U$ containing $\sigma_c$. Let us denote this analytic continuation by $A(s)$ and let $\delta > 0$ be such that $D(\sigma_c, \delta) \subset U$. Take $\sigma_0 = \sigma_c + \delta/3$. Then we have $D(\sigma_0, 2 \delta/3) \subset D(\sigma_c, \delta)$. It then follows that $A(s)$ has a power series expansion at $\sigma_0$;
\[
A(s) = \sum_{k = 0}^{\infty} \frac{A^{(k)}(\sigma_0)}{k!} (s- \sigma_0)^k \tag{1}
\] for $s \in D(\sigma_0, 2 \delta/3)$. Because $\sigma_0 > \sigma_c$ we have
\[
A^{(k)}(\sigma_0) = \alpha^{(k)}(\sigma_0) = (-1)^k\sum_{n = 1}^{\infty} a_n (\log n)^k n^{-\sigma_0}.
\] Plugging this into (1) we find that
\begin{align*} A(s) &= \sum_{k = 0}^{\infty} \frac{(-1)^k}{k!} \sum_{n = 1}^{\infty} a_n (\log n)^k n^{-\sigma_0} (s- \sigma_0)^k \\ &= \sum_{k = 0}^{\infty} \sum_{n = 1}^{\infty} \frac{1}{k!} a_n (\log n)^k n^{-\sigma_0} (\sigma_0- s)^k. \end{align*} for $s \in D(\sigma_0, 2 \delta/3)$ If $s = \sigma$ and $\sigma_c- \delta/3 < \sigma < \sigma_0$, then each summand is nonnegative and so we can interchange the order of summation to obtain
\begin{align} A(\sigma) &= \sum_{n = 1}^{\infty} a_n n^{-\sigma_0} \sum_{k = 0}^{\infty} \frac{(\log n)^k (\sigma_0- \sigma)^k}{k!} \\ &= \sum_{n = 1}^{\infty} a_n n^{-\sigma_0} e^{(\log n)(\sigma_0- \sigma)} = \sum_{n = 1}^{\infty} a_n n^{-\sigma_0} n^{\sigma_0- \sigma} = \alpha(\sigma). \end{align} This implies that $\alpha(\sigma)$ converges for $\sigma_c- \delta/3 < \sigma < \sigma_0$, a contradiction.
One can obtain an analogous result for power series (see Exercise 7 of Section 1.2 of Multiplicative Number Theory I: Classical Theory by Montgomery and Vaughan). The above result helps us to find abscissa of convergence of Dirichlet series with nonnegative coefficients. Let us illustrate this with an example. Let $d_k^*(n)$ denote the number of factorizations of $n$ as a product of $k$ positive integers each greater than $1$. Then $d_k^{*}$ is the $k^{\text{th}}$ fold convolution of $I -e$, which is the characteristic function of integers greater than $1$. Thus we have
\[
\sum_{n = 1}^{\infty} d_k^*(n) n^{-s} = (\zeta(s)- 1)^k
\] for $\sigma > 1$. This implies that the abscissa of convergence of $\sum_{n = 1}^{\infty} d_k^*(n) n^{-s}$ is $1$ as $\zeta(s)$ has a singularity at $s = 1$.