Series of holomorphic functions 
Let $a $ be a positive real number. Show that the series $\sum_{n=0}^\infty e^ {-n^az}$ converges pointwise if and only if $\mathrm {Re}(z) \gt 0$. Named the sum above $f_a $, use Morera's theorem to show that  $f_a $ is holomorphic in $\Pi =\{z:\mathrm {Re}(z)\gt 0\}$.

It is evident that if $\mathrm {Re}(z)\gt 0$, then $f _a (z) $ is finite: for $n $ big enough, $e^{-n^a}\lt e^{-2\log n}$ (since $a$ is positive), and $\sum e^{-2\log n} = \sum \frac 1 {n^2}$ converges. This doesn't change if one considers $e^{-n^ax}$ and  $ e^{-2x\log n }$, with $x $ a positive real; so the sum $\sum|e^ {-n^az}|$ converges for $z\in \Pi$. Since every term of the series is a holomorphic function of $z $, its integral over any triangle is zero; then the integral of $f_a$ over any triangle is an infinite sum of zeros, that is zero. Morera's theorem implies that $f_a $ is holomorphic in $\Pi $.

Show that $f_1$ extends to a holomorphic function over $\mathbb C\setminus 2\pi i \mathbb Z$ and study its singularities.

I don't understand this part: how can one extend $f_1$ continuously to $\Pi \cup \mathbb R i \setminus 2\pi i \mathbb Z$, if $f_1$ is not finite in any point of $\mathbb R i ?$ Instead, if $f_1$ is actually finite in $\mathbb R i \setminus 2\pi i \mathbb Z$ (and so it isn't true that $f_a $ converges pointwise only if $z\in \Pi $), I could consider the sum as $\sum (e^{-z})^n=\frac 1 {1-e^{-z}}$ ($f_1(z)$ finite implies $|e^{-zn}|\to 0)$. This would mean that there is a simple pole with residue $1$ in every $z\in 2\pi i \mathbb Z$. However I don't see how could $\sum \sin (-nx)$ and  $\sum \cos (-nx)$ converge when $x\notin 2\pi i \mathbb Z$. Can anyone clarify my ideas? Thanks in advance
 A: What you did in part 1 is correct. The series $\sum_{n=0}^\infty e^ {-n^az}$ converges if and only if $Re(z) > 0$.
Now let's go to the second part: As you are talking about an analytic continuation of $f_1$, you first have to specify a domain on which $f_1$ is holomorphic and from which you want to extend $f_1$. You did this in part 1 and found out that $f_1$ is holomorphic in $\Pi$. So you want to extend $f_1$ defined on $\Pi$. Now strictly speaking, your argument that $f_1$ cannot be continued to any point in $\mathbb R i$ because it is not finite there is wrong: $f_1$ is not defined there because now we only consider $f_1$ defined in $\Pi$. But of course, what you really wanted say is that $f_1$ is defined by $\sum_{n=0}^\infty e^ {-nz}$ in $\Pi$ and this series diverges for points in $\mathbb R i$ and so you expect that if you approach points in $\mathbb R i$, the value of $\sum_{n=0}^\infty e^ {-nz}$ will explode. 
But if you write that down carefully you note that for this expectation to hold you need to exchange a limit and a series, which requires a justification. You need the equality
$$\lim_{z \to z_0, \ z \in \Pi} \sum_{n=0}^\infty e^ {-nz} = \sum_{n=0}^\infty \lim_{z \to z_0, \ z \in \Pi} e^ {-nz} = \sum_{n=0}^\infty  e^ {-nz_0}$$
for $z_0 \in \mathbb{R} i$. The problem is that you cannot justify this step and it is indeed not true.
You can actually see that $f_1$ stays bounded near points in $\mathbb R i \setminus 2\pi i \mathbb Z$: As you correctly noted, $$f_1(z) = \sum_{n=0}^\infty e^ {-nz} = \frac 1 {1-e^{-z}}$$ for $z \in \Pi$ and it is clear that $\frac 1 {1-e^{-z}}$ stays bounded near points in $\mathbb R i \setminus 2\pi i \mathbb Z$.
Now in the end $\frac 1 {1-e^{-z}}$ defines a function which is holomorphic in $\mathbb R i \setminus 2\pi i \mathbb Z$ and it coincides with $f_1$ on $\Pi$. So it is the analytic continuation of $f_1$ you were searching for.
