I have been stuck at this problem, as my approaches have not led me to the proper result. The problem is as follows:
Prove that if $|x| <1$, then $$\frac{x}{(1-x)^2} + \frac{x^2}{(1+x^2)^2} + \frac{x^3}{(1-x^3)^2}\cdots = \frac{x}{1-x} + \frac{2x^2}{1+x^2} +\frac{3x^3}{1-x^3}\cdots$$
I tried finding the generalised term of the two sequences on either side of the equality but I couldn't go in the right approach. Any help is appreciated.
For $|x|<1$, the LHS is \begin{align*} \sum_{k=1}^{\infty}(-1)^k\frac{(-x)^k}{(1+(-x)^k)^2} &=-\sum_{k=1}^{\infty}(-1)^k\sum_{n=1}^{\infty}n(-(-x)^k)^n\\ &=-\sum_{n=1}^{\infty}(-1)^nn\sum_{k=1}^{\infty}(-1)^k((-x)^n)^k\\ &=-\sum_{n=1}^{\infty}(-1)^nn\sum_{k=1}^{\infty}(-(-x)^n)^k\\ &=\sum_{n=1}^{\infty}\frac{nx^n}{1+(-x)^n} \end{align*} which is the RHS. In the first step and in the last step we used the following identities for $|z|<1$: $$\frac{z}{(1+z)^2}=-\sum_{n=1}^{\infty}n(-z)^n \quad\mbox{and}\quad -\sum_{k=1}^{\infty}(-z)^k=\frac{z}{1+z}.$$
Note that the coefficients of this power series is the OEIS sequence A046897: the $n$ term is the sum of divisors of $n$ that are not divisible by 4.
align environment, rather than flowing horizontally.
– robjohn
Dec 16 '16 at 14:19