Quick version: I want $f'(1)$, where
$$F(z)=\frac{f(z)}{1-z}$$
with $f$ analytic at $z=1$. But when I follow a seemingly valid line of reasoning, I reach the conclusion that $f'(z)$ is not analytic at $z=1$.
Long version: I have no closed form for $f(z)$ or $F(z)$: I only have a series:
$$F(z) = \sum_{k \geq 0} a_{n} z^{n}$$
where $a_{n}$ is a sequence of positive reals (expectations of random variables, if you're curious) that's known to converge to some $a>0$, but for which I have no explicit representation. Nevertheless, since $a_{n}$ converges to a positive real number, I can conclude that $F(z)$ has a simple pole at $z=1$ (since $F$ clearly has a singularity at $1$, and a singularity of order $<1$ or $>1$ would produce coefficients that converge to $0$ or infinity respectively.)
So I can write
$$F(z) = \frac{f(z)}{1-z}$$
where f(z) is analytic at z=1. As mentioned, I want f'(1) (it's the only missing piece for the asymptotics of a sequence related to a_{n}.)
On the interval (0,1) I can then write
$$f(z)=(1-z)F(z) = (1-z) \sum_{k \geq 0} a_{k} z^{k}$$
and then transform this into
$$f(z) = a_{0} + \sum_{k \geq 1}(a_{k}-a_{k-1}) z^{k}$$
Then I differentiate term by term, yielding
$$f'(z)= \sum_{ k \geq 1} k(a_{k}-a_{k-1}) z^{k-1}$$
But this sum might diverge at $z=1$ (its convergence depends on the rate of convergence of $a_{k}$, I think.) So since this expression for $f'(z)$ is valid on $(0,1)$ , its divergence at $z=1$ implies that $f'(z)$ grows arbitrarily large near $z=1$, i.e. that $f'(z)$ has a singularity at $z=1$. And this contradicts the fact that $f(z)$ is analytic at $z=1$ by its definition.
I apologize for the vagueness of the following question, but I don't see what I'm doing wrong. I've tried considering a simpler case where $f(z)$ is a constant function, but in this case the sequence $a_{n}$ is constant, so the problematic series disappears. I've also googled "power series/analytic function with no closed form," and come up empty-handed. So any help would be uber-appreciated.
It's not true that $F(z) = \sum_{n=0}^{\infty} a_n z^n$, $|z| < 1$ must have a simple pole at $z=1$ after analytic continuation: it's not even true in general that $F$ has a meromorphic continuation to a neighborhood of $z=1$ at all.
Define $a_n$ by $a_n = \sum_{k=1}^n \frac{1}{k^2}$ and $a_0 = 0$; in other words, $F(z) = \frac{1}{1-z} \mathrm{Li}_2(z)$, where $\mathrm{Li}_2(z)$ is the polylogarithm. This has the property that $\lim_{n \rightarrow \infty} a_n = \frac{\pi^2}{6}$ is some nonzero constant, but $F$ has a logarithmic branch point at $z=1$, not a simple pole.