Using the ratio test for the series $\sum \frac1{3^n-2^n}$, I can't compute the limit $$\sum_{n=1}^{\infty} \frac{1}{3^n-2^n}$$
I know this series is convergent and using the ratio test.
But I can't conclude the proving.
$$\begin{align}\lim_{n\to\infty} \left| \frac{\frac{1}{3^{n+1}-2^{n+1}}}{ \frac{1}{3^n-2^n}}\right|
&= \lim_{n\to\infty} \left| \frac{3^n-2^n}{ 3^{n+1} - 2^{n+1}}  \right|\\
&= \lim_{n\to\infty} \left| \frac{1-(\frac{2}{3})^n}{ 3(1-2^{n+1}/3^{n+1})}\right| \end{align}$$
By using calculation, the limit is 0.
But I can't compute this limit by myself without using calculation. Any help?

You've almost got this one solved
$\lim_{n\to\infty} {(\frac{2}{3})}^n = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} .... = 0.\dot{6} \times 0.\dot{6} \times 0.\dot{6} ... = 0$
so your solution will come out to $\frac{1}{3}$ 
