Show that $\frac {1} {r-1} = \frac {1} {r+1} + \frac {2} {r^2+1} + \frac {4} {r^4+1} +\cdots$ Problem:
Show that
$$\frac {1} {r-1} = \frac {1} {r+1} + \frac {2} {r^2+1} + \frac {4} {r^4+1} +\cdots $$
for all $r > 1$, with a hint given that $\displaystyle\frac {1} {r-1} - \frac {1} {r+1} = \frac {2} {r^2-1}$.
Thoughts:
I am having difficulty seeing a connection from the hint to the problem. Any insights appreciated.
 A: First, replace $r$ by $r^{2^n}$ in your identity to get:
$$\frac{1}{r^{2^n}-1}-\frac{1}{r^{2^n}+1}=\frac{2}{r^{2^{n+1}}-1}$$
Then consider the partial sum: $s_n=\frac{1}{r+1}+\frac{2}{r^2+1}+\ldots+\frac{2^{n-1}}{r^{2^{n-1}}+1}$
You have:
$$\begin{align}\frac{1}{r-1}-s_1&=\frac{1}{r-1}-\frac{1}{r+1}=\frac{2}{r^2-1}\\\frac{1}{r-1}-s_2&=\frac{1}{r-1}-s_0-\frac{2}{r^2+1}\\
&=\frac{2}{r^2-1}-\frac{2}{r^2+1}=\frac{4}{r^4+1}\end{align}$$
And so on, in general (by induction):
$$\frac{1}{r-1}-s_n=\frac{2^n}{r^{2^n}+1}$$
Then take the limit:
$$\frac{1}{r-1}-s=\lim_{n\to\infty}\frac{2^n}{r^{2^n}+1}=\lim_{u\to\infty}\frac{u}{r^u+1}=0$$
(with substitution $u=2^n$, and using $r>1)$. So $s=\frac{1}{r-1}$
A: Note that if $r\neq1$, $\frac{n}{r^n-1}=\frac{n}{r^n+1}+\frac{2n}{r^{2n}-1}$ for any positive integer $n$.
Now, we can see taht 
$\frac{1}{r-1}=\frac{1}{r+1}+\frac{2}{r^2-1}=\frac{1}{r+1}+\frac{2}{r^2+1}+\frac{4}{r^4-1}=\cdots$ 
i.e. $\frac{1}{r-1}=\frac{2^{n+1}}{r^{2^{n+1}}-1}+\sum_{k=0}^n\frac{2^k}{r^{2^k}+1}$. 
Finally, for $r>1$, $\frac{1}{r-1}=\lim_{n\rightarrow\infty}\frac{2^{n+1}}{r^{2^{n+1}}-1}+\sum_{k=0}^n\frac{2^k}{r^{2^k}+1}=\sum_{k=0}^{\infty}\frac{2^k}{r^{2^k}+1}$. 
Good luck. 
