Proof equality of series How to prove equality :
$$\sum_{n=0}^\infty x^n \frac{1+x^{2n+2}}{(1-x^{2n+2})^2} =
  \sum_{n=0}^\infty (-1)^n \frac{x^n}{(1-x^{n+1})^2}, \: \forall x \in (-1, 1)$$
Any ideas how to start.

Start with the identity due to Euler 
$$
\prod_{m=0}^\infty\frac{1}{1-x^{2m+1}}=\prod_{m=1}^\infty(1+x^m).
$$
Taking logarithm of both sides, then expanding the logarithms into power series and reversing the order of summation we have
$$
\sum_{n=1}^\infty\frac{x^n}{n(1-x^{2n})}=\sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}{n(1-x^{n})}.
$$
Now differentiate termwise to obtain
$$\sum_{n=1}^\infty x^{n-1} \frac{1+x^{2n}}{(1-x^{2n})^2} =
  \sum_{n=1}^\infty (-1)^{n-1} \frac{x^{n-1}}{(1-x^{n})^2},
$$
which is equivalent to
$$
\sum_{n=0}^\infty x^n \frac{1+x^{2n+2}}{(1-x^{2n+2})^2} =
  \sum_{n=0}^\infty (-1)^n \frac{x^n}{(1-x^{n+1})^2}.
$$
