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.



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.

  • $\begingroup$ Wow! This is simpler than I thought! Thanks~ $\endgroup$ – GohP.iHan Dec 26 '15 at 0:19

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