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This is a rather obvious question, but my Google-fu is failing me.

Given that $F(z) = \sum_{n=1}^{\infty} a_n z^n$, can one say anything about $G(z) = \sum_{n=1}^{\infty} \frac{1}{a_n} z^n$?

In particular, I'm looking for a solution to this question using generating functions to get asymptotics for the coefficients, but I've only ever seen generating functions applied where there's an obvious way to manipulate the coefficients to match up with the recursion. It might not be possible to do this with generating functions, but I'd love to see it done.

Patrick Stevens
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  • There isn't any obvious relation between $F(z)$ and $G(z)$. For instance, if $a_n=\frac{1}{n!}$ we have that $F(z)$ is an entire function, while $G(z)$ is defined at the origin only. – Jack D'Aurizio Mar 21 '17 at 21:15
  • But of course, if you have the asymptotic behaviour of $a_n$ (through Laplace's method applied to $F(z)$, for instance) you also have the asymptotic behaviour of $\frac{1}{a_n}$. – Jack D'Aurizio Mar 21 '17 at 21:16

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