Timeline for Evaluations of three series involving binomial coefficients

6 events

when toggle format	what		by	license	comment
Nov 11 at 16:56	comment	added	CarP24		How do you prove the reciprocal cube Zete3 series? I have yet to see that one anywhere.
Feb 15, 2021 at 2:44	answer	added	Max Alekseyev		timeline score: 6
Feb 14, 2021 at 23:07	comment	added	Zhi-Wei Sun		Note also that $$\sum_{k=1}^\infty\frac1{k(-2)^k\binom{2k}k}=-\frac{\log2}3,\ \ \ \sum_{k=1}^\infty\frac1{k^2(-2)^k\binom{2k}k}=-\frac{\log^22}2$$ and $$\sum_{k=1}^\infty\frac1{k^3(-2)^k\binom{2k}k}=\frac{\log^32}6-\frac{\zeta(3)}4.$$
Feb 14, 2021 at 22:49	comment	added	Zhi-Wei Sun		It is easy to prove that $$\sum_{k=1}^\infty\frac1{k2^k\binom{3k}k}=\frac{\pi-2\log2}{10},\ \ \ \sum_{k=1}^\infty\frac1{k^22^k\binom{3k}k}=\frac{\pi^2}{24}-\frac{\log^22}2$$ and $$\sum_{k=1}^\infty\frac1{k^32^k\binom{3k}k}=\pi G+\frac{\log^22}6-\frac{\pi^2}{24}\log2-\frac{33}{16}\zeta(3).$$
Feb 14, 2021 at 16:34	history	edited	Zhi-Wei Sun	CC BY-SA 4.0	added 2 characters in body
Feb 14, 2021 at 16:22	history	asked	Zhi-Wei Sun	CC BY-SA 4.0