For $p \in [0,1]$ and $z \in \mathbb{C} - \{0\}$, I would like to evaluate (or make sense of) the complex-valued sum $F(z,p) := \sum_{ k \in \mathbb{Z}} \frac{ e^{ 2 \pi k p i }}{ z + 2 \pi i k }$. This sum is clearly not absolutely convergent, but I was wondering if something may be said about it? For instance, can something be said of the large $K$ asymptotics of $F_K(z,p) := \sum_{ |k| \leq K, k \in \mathbb{Z}} \frac{ e^{ 2 \pi k p i }}{ z + 2 \pi i k }$?
I was able to give a non-rigorous derivation of the suspicious looking $F(z,p) =^? e^{ - pz}/(1-e^{-z})$. Sketching the key steps of this computation here: with $z = x+iy$, making the denominator real, we have (formally) $F(z,p) := \bar{z} \sum_{ k \in \mathbb{Z}} \frac{ e^{ 2 \pi k p i }}{ x^2 + (2\pi k + y)^2} - \sum_{ k \in \mathbb{Z}} \frac{ 2 \pi i k e^{ 2 \pi k p i } }{ x^2 + ( 2 \pi k + y )^2} = \bar{z} G(z,p) - \frac{ \partial}{ \partial p} G(z,p)$ where $G(z,p) := \sum_{ k \in \mathbb{Z}} \frac{ e^{ 2 \pi k p i }}{ x^2 + (2\pi k + y)^2} $. Now according to the Poisson summation formula, provided $f(x)$ has exponential tails, $\sum_{ k \in \mathbb{Z}} e^{ 2 \pi i k p } f(2 \pi k)= \frac{1}{2\pi} \sum_{ k \in \mathbb{Z}} g(k + p)$ where $g(t) := \int_{ - \infty}^\infty f(u) e^{ i u t } \mathrm{d}u$. Setting $f(u) := \frac{1}{x^2 + (u+y)^2}$, we have $g(t) := \frac{ \pi e^{ - i y t } }{x} e^{ - |xt|}$. From there a calculation tells us that $G(z,p)= \frac{1}{2x} \left( \frac{ e^{ - pz}}{1 - e^{ -z}} + \frac{ e^{ - (1-p)\bar{z}} }{ 1 - e^{ - \bar{z} } } \right)$, and hence gives $F(z,p) =^? e^{ - pz}/(1-e^{-z})$.
Clearly several steps in the derivation are illegitimate or require further justification: notably the differentiation and the lack of tails for $f$.
However, is this formula for $F(z,p)$ correct in any sense? Perhaps with certain restrictions on $z$ or $p$. Ideally it should have $F(z,0) = F(z,1)$ which clearly isn't the case...