# What's the limit of coefficient ratio for a reciprocating power series?

I have a question about the coefficient in the inverse of the power series.

Assume $$f=1-\sum_{i=1}^{\infty}(ck_i)x^i,$$ where $c$ and $k_i$ are positive and $0<ck_i<1$ for any $i>0$. Define $$g=1+\sum_{i=1}^\infty t_ix^i,$$ and $$fg=1,$$ i.e., $g$ is the inverse of power series $f$.

Now I know that if $\{k_i\}$ is geometric series, i.e., $k_i=k_1^i$, then $\{t_i\}$ are also geometric series. And I remember the common ratio is $c(k_1+1)$. (If it is wrong, please point out the mistakes, thanks.)

My question is, if we don't have the condition that $\{k_i\}$ is geometric series, but we assume $$\lim_{i\rightarrow\infty}\frac{k_{i+1}}{k_i}=z_0$$ is a positive constant and less than $1$. In this case, is $$\lim_{i\rightarrow\infty}\frac{t_{i+1}}{t_i}$$ also a constant? If yes, what is it?

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Why isn't that just: $$f=1+\sum_{i=1}^{\infty}ck_ix^i$$? – Thomas Andrews Aug 30 '12 at 16:43
Sorry I made a mistake. Now I fix it. Thanks~~ – Chang Aug 31 '12 at 3:15
Bah, got a bit confused reading this; maybe edit the question to say you're reciprocating instead of inverting? – J. M. Sep 1 '12 at 7:44
I changed the title followed your advise~~Thanks~~ – Chang Sep 1 '12 at 7:49
When one speaks about inverse power series then one has the case $f(0)=0$, $f'(0)\ne0$ in mind. Because of the $1$ in front your series for $f$ and $g$ cannot be composed in a finitary way. The case of geometric series (= Moebius functions) is very special. – Christian Blatter Sep 1 '12 at 7:59

As I explained in this answer to an earlier question of yours, $g$ will have a positive radius of convergence, determined by the location of the complex zero of $f$ closest to the origin (provided such a zero exists within the disk where the series for $f$ converges). This radius of convergence is in general quite unrelated to the radius of convergence $z_0$ of $f$ itself.

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