Problem source

Given that $b>a>0$, find, by using the binomial theorem, coefficients $c_{m}$ ($m=0,1,2,\ldots$) such that
\[
\frac{1}{\left(1-ax\right)\left(1-bx\right)}=c_{0}+c_{1}x+c_{2}x^{2}+\ldots+c_{m}x^{m}+\cdots
\]
for $b\left|x\right|<1$. 
Show that 
\[
c_{m}^{2}=\frac{a^{2m+2}-2(ab)^{m+1}+b^{2m+2}}{(a-b)^{2}}\,.
\]
Hence, or otherwise, show that
\[
c_{0}^{2}+c_{1}^{2}x+c_{2}^{2}x^{2}+\cdots+c_{m}^{2}x^{m}+\cdots=\frac{1+abx}{\left(1-abx\right)\left(1-a^{2}x\right)\left(1-b^{2}x\right)}\,,
\]
for $x$ in a suitable interval which you should determine.

Solution source

\begin{align*}
\frac{1}{(1-ax)(1-bx)} &=\frac{1}{b-a} \l \frac{b}{1-bx}-\frac{a}{1-ax}\r \\
&= \frac{1}{b-a} \l \sum_{k=0}^{\infty} b(bx)^k-\sum_{k=0}^{\infty} a(ax)^k \r \\
&= \frac{1}{b-a} \sum_{k=0}^{\infty} \l b^{k+1} - a^{k+1} \r x^k
\end{align*}

Therefore $\displaystyle c_m = \frac{b^{k+1}-a^{k+1}}{b-a}$.

\begin{align*}
c_m^2 &= \frac{(b^{m+1}-a^{m+1})^2}{(b-a)^2} \\
&= \frac{a^{2m+2} - 2(ab)^{m+1} + b^{2m+2}}{(b-a)^2}
\end{align*}

\begin{align*}
\sum_{m=0}^{\infty} c_m x^m &= \sum_{m=0}^{\infty}  \l \frac{a^{2m+2} - 2(ab)^{m+1} + b^{2m+2}}{(b-a)^2} \r x^m  \\
&= \frac{1}{(b-a)^2} \l \sum_{m=0}^{\infty} a^{2m+2} x^m-2\sum_{m=0}^{\infty} (ab)^{m+1} x^m+\sum_{m=0}^{\infty} b^{2m+2} x^m \r \\
&= \frac{1}{(b-a)^2} \l a^2\sum_{m=0}^{\infty} a^{2m} x^m-2ab\sum_{m=0}^{\infty} (ab)^{m} x^m+b^2\sum_{m=0}^{\infty} b^{2m} x^m \r \\
&= \frac{1}{(b-a)^2} \l \frac{a^2}{1-a^2x^2} - \frac{2ab}{1-abx} + \frac{b^2}{1-b^2x^2}\r \\
&= \frac{1+ab}{(1-a^2x)(1-abx)(1-b^2x)}
\end{align*}

Where geometric series will converge if $|a^2x| < 1, |b^2x| < 1, |abx| < 1$, ie $|x| < \min (\frac{1}{a^2}, \frac{1}{b^2} )$