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!

  • $\begingroup$ Now I know how to prove the identity. $\endgroup$ May 1, 2022 at 11:58
  • $\begingroup$ It is clear that the same method will work for this. $\endgroup$ May 1, 2022 at 12:00

1 Answer 1


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 $$ LHS=-\frac{\log \left(2-\sqrt{3}\right)}{4 \sqrt{3}}\neq \frac{\pi }{12 \sqrt{3}}=RHS. $$

  • $\begingroup$ I have just reedited my posting and corrected the identity form. $\endgroup$ May 1, 2022 at 11:22
  • $\begingroup$ @Zhi-WeiSun does the method in this answer work? $\endgroup$ May 1, 2022 at 19:29
  • $\begingroup$ I use another approach to prove the identity. $\endgroup$ May 1, 2022 at 21:14
  • $\begingroup$ @Utas You are welcome to modify your answer to prove the identity fullly and rigoriously. $\endgroup$ May 2, 2022 at 2:22

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