I know Landau's theorem says that if we have a Dirichlet series with nonnegative coefficents with abscissa $a$, then it is impossible to find an analytic extension of our function which is defined also on $a$ and a neighborhood of $a$.
Does it follow that the Dirichlet series tends to infinity as $s$ tends to $a$? I know that if we already had an analytic function defined on a neighborhood of $a$, then we can extend it to $a$ if the limit is bounded, but apriori why would we have that?
Thanks.
It does not follow that $f(s)$ tends to infinity as $s$ tends to $a$. It is possible that the series converges (absolutely) on the line $\operatorname{Re} s = a$.
As an example of such a situation, consider the series
$$f(s) = \sum_{n = 1}^{\infty} \frac{1}{(1 + \log n)^2\cdot n^s}.$$
Its abscissa of convergence is $1$, and the series converges absolutely on the line $\operatorname{Re} s = 1$, as can be seen for example using the integral test.