Recently I formulated the following curious conjecture based on my computation.
Conjecture. For all $|x|>1$, we have the identity $$\sum_{k=0}^\infty\frac{\sum_{j=0}^{k}\binom{2k+1}{2j}(1-x)^jx^{k-j}}{(2k+1)(2x-1)^{2k+1}}=\frac1{2}\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)x^{k+1}}.\tag{1}$$
QUESTION. Is the conjecture true? Can one provide a proof of $(1)$?
I don't think the problem is very difficult. Your comments are welcome!
Let $Q_{k}(x)=\sum_{k=0}^{2k}\binom{2k+1}{2j}(1-x)^{j}x^{k-j}$. One can check that $Q_{0}(x)=1, Q_{1}(x)=3-2x$ and for $k\geq 2$ we have the following recurrence: $$ Q_{k}(x)=2Q_{k-1}(x)-(2x-1)^{2}Q_{k-2}(x). $$ Using standard methods one can find exact expression for $Q_{k}(x)$ in the form $Q_{k}(x)=P_{1}(x)r_{1}(x)^{k}+P_{2}(x)r_{2}(x)^{k}$, where $$ r_{1}(x)=1-2\sqrt{x-x^2},\quad r_{2}(x)=1+\sqrt{x-x^2} $$ and $$ P_{1}(x)=\frac{x-1+\sqrt{(1-x) x}}{2 \sqrt{(1-x) x}},\quad P_{2}(x)=\frac{1-x+\sqrt{(1-x) x}}{2 \sqrt{(1-x) x}}. $$ Using the identity $$ \sum_{k=0}^{\infty}\frac{u^{k}}{(2k+1)v^{2k+1}}=\frac{\tanh ^{-1}\left(\frac{\sqrt{u}}{v}\right)}{\sqrt{u}} $$ (with appropriate values of $u, v$) one can obtain closed form expression for LHS (quite complicated) and RHS. However, it seems that the formula is still incorrect. Take $x=3$. Then $$ LHS=-\frac{\log \left(2-\sqrt{3}\right)}{4 \sqrt{3}}\neq \frac{\pi }{12 \sqrt{3}}=RHS. $$
