Holomorphy of a series of complex functions

Question

I have to discuss the holomorphy of the function $$f (z)=\sum_{n=1}^\infty \frac {e^{nz}} {z-n}, $$ and study its singular points. I would say that in the part of the complex with $\mathrm {Re} (z) \lt 0$, $f $ is holomorphic, since the sum converges; instead, when $\mathrm {Re} (z) \gt 0$, the terms of the series go to infinity as $n $ increases, so the sum clearly doesn't converge. My doubt is for the points with real part $0$: in the points $2\pi i k $, with $k $ integer, the sum doesn't converge because the numerator is $1$. In the other points of the imaginary axis the sum of the $|\frac {e^{nz}} {z-n}|$ doesn't converge, but the sum above could converge anyway. However, does it make sense to study the convergence in that points? I mean, even if the convergence was verified, the function wouldn't be holomorphic in a neighbourhood of a point on the imaginary axis; so this function can only be holomorphic in the open set with $\mathrm {Re} (z) \lt 0$. However suppose that the problem asked me to determine if the same function, defined in $\mathrm {Re} (z) \le 0$, was extendable to a holomorphic function in $\mathbb C \setminus \{2\pi i k \} $, $k $ integer. Now would make sense to study the convergence in the point of the imaginary axis, so in general how do you proceed to do it? Thank you

Answer 1

Let $z=ix$, $x$ real. Then $|\frac{e^{inx}}{ix-n}-\frac{e^{inx}}{-n}|=O(\frac{1}{n^2}), n \ge 1$ and $x$ in a compact (hence bounded) subset of the reals, so for example fixing it, hence the series has the same convergence as the well known series $\sum{\frac{\cos nx}{n}}, \sum{\frac{\sin nx}{n}}$ which converge precisely when $ x \ne 2k\pi$ by the usual methods (Abel summation etc).

However, this has little to do with extending the function holomorphically beyond the imaginary line - we know that $\Re f$ is unbounded near $2k\pi i$ so we definitely cannot extend there.

Edit - later I realized that I was hasty here and actually the series of second derivatives is not unbounded near the imaginary line and to the left - the behavior on the imaginary line of the series is irrelevant and the function can be extended to the right plane. Also the unboundness in the left plane near the critical values needs a better argument, though it is still provable.