I was working on a problem, and I had made the assumption that given a Dirichlet series $$ L(s,f)=\sum_{n\geq 1}\frac{f(n)}{n^s} $$ If I have some $\sigma\in\mathbb{C}$ such that $L(\sigma,f)$ diverges, then I had assumed that $L(\sigma,f)\rightarrow\infty$. However, after thinking about it, I realized that this is false as I had considered the case of $f(n)=(-1)^n$ as then $$ L(-1,f)=\sum_{n\geq 1}(-1)^nn $$ and in this case, we have that this series does not diverge to infinity (thinking of it as a series of real numbers, but I guess if you think of it on the Riemann sphere it does approach infinity). Thus, I then thought about if it is always the case that if $L(\sigma,f)$ diverges, then $\vert L(\sigma,f)\vert\rightarrow\infty$. However, I then realized that this is also too much to ask for since with my above example we have that $$ L(0,f)=\sum_{n\geq 1}(-1)^n $$ which diverges, but does not go off to infinity.
Thus, my question is whether or not there is some criteria on $\sigma$ or $f(n)$ that says if $L(\sigma,f)$ diverges then $\vert L(\sigma,f)\vert\rightarrow\infty$ or even $L(\sigma,f)\rightarrow\infty$? Also, if $\vert L(\sigma,f)\vert\not\rightarrow\infty$, can we conclude that there is a removable singularity at $s=\sigma$.
There is some confusion between the Dirichlet series and its analytic continuation (which doesn't have to exist..)
For the Dirichlet series things are relatively simple. If $A_{s_0}(m)=\sum_{n\le m} a_n n^{-s_0}$ is bounded then $$\sum_{n\ge 1}^\infty a_n n^{-s} = \lim_{N\to \infty} A_{s_0}(N) N^{s_0-s} +\sum_{m=1}^{N-1} A_{s_0}(m) (m^{s_0-s}-(m+1)^{s_0-s})$$ converges for all $\Re(s) > \Re(s_0)$.
So if $\Re(s_0)$ is smaller than the abscissa of convergence then $$\sup_m|A_{s_0}(m)|=\infty$$