$$\sum_{n=0}^\infty\frac{1}{3^{2^n}-3^{-2^n}}=\frac{1}{2}$$
I assumed this is a geometric series, but im not 100% sure about it. I checked if its convergent anyway, and it is, but other than that I cant prove it converges to $\frac{1}{2}$. Limit when n approaches infinity of $x_n$ is 0.
To see that this sum is telescopic, rewrite the general term as
$$\frac {3^{2^n}} {3^{2^{n+1}} - 1} = \frac 1 {3^{2^n} - 1} - \frac 1 {3^{2^{n+1}} - 1}$$
Another way is to simply observe the partial sums
\begin{alignat}{2} \frac38 &= \ \ \ \frac38 & & = \frac12-\frac18\\ \frac38 + \frac 9 {80} &= \ \ \frac {39}{80} & &= \frac12-\frac1{80}\\ \frac38 + \frac 9 {80} + \frac {81} {6560} &= \frac {3279}{6560}& &= \frac12-\frac1{6560} \end{alignat}
and try to prove this observation by induction.