Find the sum $\sum_{n=1}^{\infty} \frac{3^n}{5^n-2^{2n}}$. Can someone help me with this sum $\sum_{n=1}^{\infty} \frac{3^n}{5^n-2^{2n}}$. I can't find $S_n$
 A: $$\begin{eqnarray*}S&=&\sum_{n\geq 1}\frac{3^n}{5^n-4^n}=\sum_{n\geq 1}\left(\frac{3}{5}\right)^n\frac{1}{1-\left(\frac{4}{5}\right)^n}=\sum_{n\geq 1}\sum_{k\geq 0}\left(\frac{3}{5}\right)^n\left(\frac{4}{5}\right)^{kn}=\sum_{k\geq 0}\frac{3\cdot 4^k}{5^{k+1}-3\cdot 4^{k}}\\&=&\sum_{m\geq 1}\frac{\frac{3}{4}4^m}{5^m-\frac{3}{4}4^m}\end{eqnarray*} $$
is actually a series deceleration, but there are ways to improve the convergence speed of the LHS.
For instance
$$ S = 3+\sum_{n\geq 2}\frac{3^{n}}{5^n-4^n}=3+\sum_{n\geq 1}\frac{3^{n+1}}{5^{n+1}-4^{n+1}} $$
where the last series is pretty close to $\frac{3}{5}S$, leading to
$$ \frac{2}{5}S = 3 + \sum_{n\geq 1}\left(\frac{3^{n+1}}{5^{n+1}-4^{n+1}}-\frac{3}{5}\cdot\frac{3^{n}}{5^{n}-4^{n}}\right) $$
and
$$ S = \frac{15}{2}-\frac{3}{2}\sum_{n\geq 1}\frac{3^{n}4^n}{(5^{n+1}-4^{n+1})(5^{n}-4^n)}.$$
The same trick can be applied to the last series, whose main term roughly behaves like $\left(\frac{12}{25}\right)^n$.
$$ S = \frac{95}{26}+\frac{162}{13}\sum_{n\geq 1}\frac{3^n 4^{2n}}{(5^{n+2}-4^{n+2})(5^{n+1}-4^{n+1})(5^n-4^n)} $$
After a few steps we get $S\approx 4.92476079$, but I highly doubt there is a nice closed form fo $S$. After all this is pretty much the acceleration method employed by Shingo Takeuchi (and reminiscent of Gosper's works in the seventies) for proving the trascendence of some series like $\sum_{k\geq 1}\frac{1}{2^k-1}$.
A: You can write $$\frac{3^n}{5^n-2^n} = \frac{(3/5)^n}{1 - (2/5)^n} = \sum_{k=0}^\infty (3/5)^n (2/5)^{kn}$$
so your sum is
$$ \eqalign{ \sum_{n=1}^\infty \sum_{k=0}^\infty (3/5)^n (2/5)^{kn}
&= \sum_{k=0}^\infty \sum_{n=1}^\infty (3/5)^n (2/5)^{kn}\cr
&= \sum_{k=0}^\infty \frac{3 \cdot 2^k}{5^{1+k}-3 \cdot 2^k}}$$
But that's no closer to a "closed form" than the original.
