I'm working with infinite geometric-like sums of the form $\sum_{n=0}^\infty e^{f(n)}$. Let's call this a transformation $X$ on $f$. I.e;
$$X\{f\}=\sum_{n=0}^\infty e^{f(n)}$$
Essentially, with the view that this is a transformation of $f$, I'm hoping it is possible to represent this transformation in another way. Perhaps as an integral. It's hard to describe exactly what I'm asking, so do forgive my lack of proper terminology.
Let me share my thought process with two examples, in hopes that somebody may be able to formalize what I'm asking.
Take the example $f(n)=an+b$. Thus, the transformation is trivial through the geometric series formula.
$$X\{an+b\}=\frac{e^{b}}{1-e^{a}}$$
I notice that this can equivalently be written as $$\frac{e^{f(0)}}{1-e^{f(1)-f(0)}}$$
Of course, this generalization is rediculous, and does not hold for nearly any other $f$, for instance take $f(n)=an^2$. This is less trivial, but thanks to WolframAlpha, it can be expressed using the Jacobi Theta functions:
$$X\{an^2\}=\frac{1}{2}(1+\vartheta_3(0,e^a))$$
Which we can pseudo-generalize as
$$\frac{1}{2}(1+\vartheta_3(0,e^{f(1)}))$$
Again, such a generalization is silly. However, I hope it outlines that I'm trying to find some representation of this transformation in terms of $f$, besides the infinite sum. I.e; a representation that shows how we get from the input function to the form of the result, which would answer what the relationship is between $an+b$ and $\frac{e^{b}}{1-e^{a}}$, or $an^2$ and the Jacobi Theta functions. Undoubtedly, such a representation will not be have closed form, but it would be really nice to be able to manipulate this infinite sum into an integral or something of the like.
(I should point out that this is all, of course, assuming the transformation converges. If it doesn't, it would be nice to assume the result yields something analogous to analytic continuation.)
Update:
It appears that the Abel-Plana formula is what I was looking for. Adapted for my "transformation":
$$\sum_{n=0}^\infty e^{f(n)}=X\{f\}=\int_0^\infty e^{f(x)}dx+\frac{1}{2}e^{f(0)}+i\int_0^\infty\frac{e^{f(it)}-e^{f(-it)}}{e^{2\pi t}-1}dt$$