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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 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?

I don't know much about this. Can you help me? Thanks in advance.

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    Hello @Marc van Leeuwen, thank you for pointing out this. I am reading relative information these days. Yesterday I saw the radius of convergence, and found it can be expressed by the limit. But I didn't realize that the limitation is stronger. Thanks so much~~And I need to clarify that I hope to prove the limit of $t_i$ ratio is exist. And I will correct the question.2012-09-01

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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.