<s>[INST] 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
 [/INST] You may continue as:
$$\frac{2^n}{(2^n-1)(2^n+1)} =\frac{(2^n+1)-1}{(2^n-1)(2^n+1)} =\frac 1{2^n-1} - \frac 1{(2^n-1)(2^n+1)}$$
Where $n = 2^r$. Now write the sum as:
$$\left(\frac 11 - \frac 13\right)  + \left(\frac 13 - \frac 1{15}\right) + ... + \left(\frac 1{2^n-1} - \frac 1{(2^n-1)(2^n+1)}\right)$$
After cancelling the terms you are left with:
$$1 - \frac 1{(2^n-1)(2^n+1)}$$
When $n$ tends to infinity, the expression becomes $1$.
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