Algebraic Expression Hiding In Plain Sight

Since we're finding the infinite sum, we should always check whether this sum telescopes or not. And because the entire expression, $\dfrac{2^n}{(2^n-1)(2^{n+1} - 1) }=\dfrac{2^n}{(2^n-1)(2\cdot 2^{n} - 1) }$ can be expressed in terms of the function of $2^n$ , this motivates me to replace $m = 2^n$ so that we have an algebraic expression, $\dfrac m{(m-1)(2m-1)}$ .

And we because we want to find whether this sum telescopes or not, the next move should be partial fractions , and because all the factors are linear factors, this further motivates me to use cover up rule , and so we have

$\dfrac m{(m-1)(2m-1)} \equiv \dfrac1{m-1} - \dfrac1{2m-1}$

Substituting back $m = 2^m$ gives us $\dfrac{2^n}{(2^n-1)(2^{n+1} - 1) } = \dfrac1{2^n-1} - \dfrac1{2\cdot 2^n - 1} = \dfrac1{2^n-1} - \dfrac1{ 2^{n+1} - 1} .$

And let's check if this series does telescope or not. If we let $a_n = \dfrac1{2^n - 1}$ , then $a_{n+1} = \dfrac1{2^{n+1} - 1}$ , hence the series in question becomes

$\begin{aligned} && \displaystyle\sum_{n=1}^{\infty}\dfrac{2^{n}}{(2^{n}-1)(2^{n+1}-1)} \\ &=& \displaystyle\sum_{n=1}^{\infty} ( a_n - a_{n+1} ) \\ &=& \displaystyle \lim_{N\to\infty} \sum_{n=1}^N ( a_n - a_{n+1}) \\ &=& \lim_{N\to\infty} \left [ (a_1 -a_2) + (a_2 - a_3) + (a_3 - a_4) + \cdots + ( a_{n-1} - a_n ) + ( a_n - a_{n+1} ) \right ] \\ &=& \lim_{N\to\infty} \left [ (a_1 -\xcancel{a_2}) + (\xcancel{a_2} - \xcancel{a_3}) + (\xcancel{a_3} - \xcancel{a_4}) + \cdots + ( \xcancel{a_{n-1}} - \xcancel{a_n} ) + ( \xcancel{a_n} - a_{n+1} ) \right ] \qquad \text{ It telescopes!!} \\ &=& \lim_{N\to\infty} (a_1 - a_{n+1} ) \\ &=& \lim_{N\to\infty} \left (1 - \dfrac1{2^{n+1} - 1} \right ) \\ &=& 1- 0 = \boxed1\\ \end{aligned}$