Coefficients of Quotient of two power series So I have two functions $f$ and $g$ which are holomorphic on some disk $D(a,r)$ and such that $g$ is never zero on that disk.
Where we can represent the two function as power series given by
$f(z)=\sum\limits_{n=0}^\infty a_n(z-a)^n$ and  $g(z)=\sum\limits_{m=0}^\infty b_m(z-a)^m$ 
Let
$f_N=\sum\limits_{n=0}^N a_n(z-a)^n$ and $g_N=\sum\limits_{m=0}^N b_m(z-a)^m$ then by carrying out the long division of the two polynomials we get that
$$f_N=p_Ng_N+r_N$$ where $p_N$ and $r_N$ are two polynomials and $deg(r_N)$ is at most $N-1$
I want to show that for each integer $m$, the coefficient $c_m$ of $(z-a)^m$ is the same in all the polynomials $p_N$ for $N>m$, and that will be the coefficients on $z^m$ in the Taylor series for $f/g$.
I'm not sure how to do that. 
Perhaps someone could give me a hint or tell me if I'm going about this the wrong way.
What I have so far is that I know that since $f$ and $g$ are holomorphic on $D(a,r)$ and since $g$ is never zero on $D(a,r)$, $f/g$ will be holomorphic on $D(a,r)$ also. Thus $f/g$ has a power series representation on $|z-a|<r$.
I also know that for every positive integer $m$ and every $s$ such that $0<s<r$,
\begin{align*}c_m&=\frac{1}{2\pi i}\int_{|w-a|=s}\frac{f(w)}{g(w)(w-a)^{m+1}}dw\\
&=\lim\limits_{N\to\infty}\frac{1}{2\pi i}\int_{|w-a|=s}\frac{f_N(w)}{g_N(w)(w-a)^{m+1}}dw\\
&=\lim\limits_{N\to\infty}\frac{1}{2\pi i}\int_{|w-a|=s}\frac{p_N(w)g_N(w)+r_N(w)}{g_N(w)(w-a)^{m+1}}dw\\
&=\lim\limits_{N\to\infty}\frac{1}{2\pi i}\int_{|w-a|=s}\left[\frac{p_N(w)}{(w-a)^{m+1}}+\frac{r_N(w)}{g_N(w)(w-a)^{m+1}}\right]dw\\
&=\lim\limits_{N\to\infty}\frac{1}{2\pi i}\int_{|w-a|=s}\frac{p_N(w)}{(w-a)^{m+1}}dw+\lim\limits_{N\to\infty}\frac{1}{2\pi i}\int_{|w-a|=s}\frac{r_N(w)}{g_N(w)(w-a)^{m+1}}dw\\
\end{align*}
I think that the first integral gives the coefficients of the polynomial $p$, but I'm not sure what happens with the second integral since $deg(g_N)>deg(r_N)$.
 A: Suppose $f(z)=\sum\limits_{n=0}^\infty a_n(z-a)^n$ and $g(z)=\sum\limits_{m=0}^\infty b_m(z-a)^m$. Then
You can compute the quotient of the infinite series by writing the quotient with a denominator of the form $1-\sum\limits_{m=1}^\infty -b_m(z-a)^m$. Dividing by this denominator is the same as multiplying by a geometric series.
$$\dfrac{\sum\limits_{n=0}^\infty a_n(z-a)^n}{\sum\limits_{m=0}^\infty b_m(z-a)^m}=
\dfrac{\sum\limits_{n=0}^\infty a_n(z-a)^\infty}{b_0-\left(\sum\limits_{m=1}^\infty -b_m(z-a)^m\right)}\\
=\frac{1}{b_0}\sum\limits_{n=0}^\infty a_n(z-a)^n\left(1+ \left(\sum\limits_{m=1}^\infty \frac{-b_m}{b_0}(z-a)^m\right)+\left(\sum\limits_{m=1}^\infty \frac{-b_m}{b_0}(z-a)^m\right)^2+\cdots\right)$$
An example might help:
$$\dfrac{e^z}
{\frac{2}{2-z^2}}=\dfrac{\sum\limits_{n=0}^\infty \frac{z^n}{n!}}
{\sum\limits_{m=0}^\infty \frac{z^{2m}}{2^m}}
=
\dfrac{\sum\limits_{n=0}^\infty \frac{z^n}{n!}}{{2^0}-\left(\sum\limits_{m=1}^\infty -\frac{z^{2m}}{2^m}\right)}\\
=\frac{1}{2^0}\sum\limits_{n=0}^\infty \frac{z^n}{n!}\left(1 +\left(\sum\limits_{m=1}^\infty -\frac{z^{2m}}{2^{m+0}}\right)+\left(\sum\limits_{m=1}^\infty -\frac{z^{2m}}{2^{m+0}}\right)^2+\cdots\right)
\\\left(1+z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}\cdots\right)\left(1 -\left(\frac{z^2}{2}-\frac{z^4}{4}-\frac{z^6}{8}\cdots\right)+\left(\frac{z^2}{2}-\frac{z^4}{4}-\frac{z^6}{8}\cdots\right)^2+\cdots\right)
\\=1+z+\frac{z^2}{2}-\frac{z^2}{2}+\frac{z^3}{6}-\frac{z^3}{2}\cdots=1+z-\frac{z^3}{3}\cdots$$
You can write the coefficients of the quotient series in “closed form” (there will be finite sums in each term) if you google division of power series, but it is nice to see how to do it by hand term by term, as shown here.
