# Prove (or disprove) that $\sum_{n=1}^\infty \frac{4(-1)^n}{1-4n^2} x^n = \frac{2(x+1) \tan^{-1}(\sqrt x)}{\sqrt x} - 2$ for $0<x\leq1$

Just like title said, for $0 <x\leq1$, prove/disprove:

$$\displaystyle \sum_{n=1}^\infty \dfrac{4(-1)^n}{1-4n^2} \cdot x^n \stackrel{?}{=} \dfrac{2(x+1) \tan^{-1}(\sqrt x)}{\sqrt x} - 2$$

I got this equation from Claude Leibovici. It's true for $n=1$ as shown by Ron Gordon.

I think it's feasible to show that it's true from the RHS by converting the expression into a Maclaurin Series, but I was curious if there's a way to solve this problem with reverse engineering it.

MERRY XMAS!!!

By integrating $$\frac{1}{1+x^2}=1-x^2+x^4-x^6+\cdot\cdot\cdot$$ one has $$\arctan x=\Sigma_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1}.$$ After some manipulation $$2\Big[\frac{x+1}{\sqrt x}\arctan(\sqrt x)-1\Big]=2x+2\Sigma_{n=1}^{\infty}(-1)^n\frac{(x+1)x^n}{2n+1}.$$Now collect terms with same powers of $x$ by regrouping coefficients. Say for example the coefficient of $x^k$ would be $$C_k=2\Big[\frac{(-1)^k}{2k+1}+\frac{(-1)^{k-1}}{2k-1}\Big].$$ It follows from here.