Infinite series problem

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.

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 $1$.

Note first that $$ \frac{2^n}{2^{2n}-1}=2^{-n}\frac{1}{1-2^{-2n}}=\sum_{k:\text{odd},k\in\Bbb N} 2^{-nk}. $$ If we sum over $n=2^j$, $j\ge 0$, we have $$ S=\sum_{j=0}^\infty \frac{2^{2^j}}{2^{2^{j+1}}-1}=\sum_{j=0}^\infty\sum_{k:\text{odd},k\in\Bbb N} 2^{-2^j\cdot k}=\sum_{l=1}^\infty 2^{-l}=1 $$ since every $l\ge 1$ has a unique representation $l=2^j\cdot k$ for some $j\ge 0$ and odd $k\in \Bbb N$.