Infinite series problem 
The sum of $$\frac{2}{4-1}+\frac{2^2}{4^2-1}+\frac{2^4}{4^4-1}+\cdots \cdots $$

Try: write it as  $$S = \sum^{\infty}_{r=0}\frac{2^{2^{r}}}{2^{2^{r+1}}-1}=\sum^{\infty}_{r=0}\frac{2^{2^r}-1+1}{(2^{2^r}-1)(2^{2^r}+1)}$$
d not know how to solve from here, could some help me 
to solve it, Thanks

You may prove by induction that $$\sum_{r=0}^n\frac{2^{2^r}}{2^{2^{r+1}}-1}=1-\frac{1}{2^{2^{n+1}}-1}.$$(If yoiu didn't spot this conjecture at first, you will after calculating the first few partial sums.) Indeed the claim is correct if $n=0$, and if it holds for $n=k$ then $$\sum_{r=0}^{k+1}\frac{2^{2^r}}{2^{2^{r+1}}-1}=1-\frac{1}{2^{2^{k+1}}-1}+\frac{2^{2^{k+1}}}{2^{2^{k+2}}-1}=1-\frac{1}{2^{2^{k+2}}-1}$$as required, where the final calculation is the $a=2^{2^{k+1}}$ special case of $$1-\frac{1}{a-1}+\frac{a}{a^2-1}=1-\frac{1}{a^2-1}.$$You may wish to rewrite this argument as the computation of a telescoping series.
