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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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    Why isn't that just: $$f=1+\sum_{i=1}^{\infty}ck_ix^i$$?2012-08-30
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    Sorry I made a mistake. Now I fix it. Thanks~~2012-08-31
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    Bah, got a bit confused reading this; maybe edit the question to say you're *reciprocating* instead of inverting?2012-09-01
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    I changed the title followed your advise~~Thanks~~2012-09-01
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    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.2012-09-01
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    The fact that $\lim_{i\rightarrow\infty}\frac{k_{i+1}}{k_i}=z_0\neq0$ _implies_ that the radius of convergence is $\frac1{z_0}$, but is a stronger statement. For instance your coefficients could be alternately zero (or extremely small) and nozero, and your limit would not exist, but you can still have the given radius of convergence. Please be specific if you really need the limit to exist (for the $t_i$); the radius of convergence part is much easier to handle.2012-09-01
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    Hello @ChristianBlatter. I really appreciate your correct, which is much helpful to me.2012-09-01
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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.