How to show that $\sum_{n=1}^{\infty}{2^n\over (2^n-1)(2^{n+1}-1)(2^{n+2}-1)}={1\over 9}?$ How to show that
$${2\over (2-1)(2^2-1)(2^3-1)}+{2^2\over (2^2-1)(2^3-1)(2^4-1)}+{2^3\over (2^3-1)(2^4-1)(2^5-1)}+\cdots={1\over 9}?\tag1$$
We may rewrite $(1)$ as
$$\sum_{n=1}^{\infty}{2^n\over (2^n-1)(2^{n+1}-1)(2^{n+2}-1)}={1\over 9}\tag2$$
$${2^n\over (2^n-1)(2^{n+1}-1)(2^{n+2}-1)}={A\over 2^n-1}+{B\over 2^{n+1}-1}+
{C\over 2^{n+2}-1}\tag3$$
$${2^n\over (2^n-1)(2^{n+1}-1)(2^{n+2}-1)}={1\over 3(2^n-1)}-{1\over 2^{n+1}-1}+
{2\over 3(2^{n+2}-1)}\tag4$$
$${1\over 3}\sum_{n=1}^{\infty}{1\over 2^n-1}-\sum_{n=1}^{\infty}{1\over 2^{n+1}-1}+{2\over 3}\sum_{n=1}^{\infty}{1\over 2^{n+2}-1}={1\over 9}\tag5$$
 A: Well done so far.  Now shift the indices on the last two sums:
$${1\over 3}\sum_{n=1}^{\infty}{1\over 2^n-1}-\sum_{n=1}^{\infty}{1\over 2^{n+1}-1}+{2\over 3}\sum_{n=1}^{\infty}{1\over 2^{n+2}-1}=\\
{1\over 3}\sum_{n=1}^{\infty}{1\over 2^n-1}-\sum_{n=2}^{\infty}{1\over 2^{n}-1}+{2\over 3}\sum_{n=3}^{\infty}{1\over 2^{n}-1}$$
and note that all the terms with $n \ge 3$ add to zero, so we just keep the first few terms:
$$=\frac 13 \cdot \frac 1{2-1}+\frac 13\cdot \frac 1{4-1}-\frac 1{4-1}=\frac 19$$
A: You are almost done. Note that the three series on the LHS of your last line are convergent and they can be written as
$${1\over 3}\sum_{n=1}^{\infty}{1\over 2^n-1}-\sum_{n=2}^{\infty}{1\over 2^{n}-1}+{2\over 3}\sum_{n=3}^{\infty}{1\over 2^{n}-1}$$
which is equal to
$$
{1\over 3}\sum_{n=1}^{2}{1\over 2^n-1}-\sum_{n=2}^{2}{1\over 2^{n}-1}+\left(\frac{1}{3}-1+{2\over 3}\right)\sum_{n=3}^{\infty}{1\over 2^{n}-1}=\frac{1}{9}.$$
A: Hint: We can express
$$\frac{1}{2^n-1}=\sum_{k=1}^{\infty} \frac{1}{2^{kn}}.$$
