# Evaluations of three series involving binomial coefficients

Question. How to prove the following three identities? \begin{align}\sum_{k=1}^\infty\frac1{k(-2)^k\binom{2k}k}\left(\frac1{k+1}+\ldots+\frac1{2k}\right)=\frac{\log^22}3-\frac{\pi^2}{36},\tag{1} \end{align} \begin{align}\sum_{k=1}^\infty\frac1{k2^k\binom{3k}k}\left(\frac1{k+1}+\ldots+\frac1{2k}\right) =\frac{3}{10}\log^22+\frac{\pi}{20}\log2-\frac{\pi^2}{60},\tag{2} \end{align}\begin{align}\sum_{k=1}^\infty\frac1{k^22^k\binom{3k}k}\left(\frac1{k+1}+\ldots+\frac1{2k}\right) =-\frac{\pi G}2+\frac{33}{32}\zeta(3)+\frac{\pi^2}{24}\log2,\tag{3} \end{align} where $$G$$ denotes the Catalan constant $$\sum_{k=0}^\infty(-1)^k/{(2k+1)^2}$$.

Remark. Motivated by my study of congruences, in 2014 I tried to evaluate the three series in $$(1)$$-$$(3)$$, and this led me to discover $$(1)$$-$$(3)$$ which can be easily checked numerically via Mathematica. But I'm unable to prove the above three identities. Also, Mathematica could not evaluate the three series. For more backgrounds of this topic, you may visit http://maths.nju.edu.cn/~zwsun/165s.pdf.

• 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 $$\sum_{k=1}^\infty\frac1{k^32^k\binom{3k}k}=\pi G+\frac{\log^22}6-\frac{\pi^2}{24}\log2-\frac{33}{16}\zeta(3).$$ Commented Feb 14, 2021 at 22:49
• 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 $$\sum_{k=1}^\infty\frac1{k^3(-2)^k\binom{2k}k}=\frac{\log^32}6-\frac{\zeta(3)}4.$$ Commented Feb 14, 2021 at 23:07
We have $$\frac1{k\binom{2k}{k}}=\frac12 B(k,k)=\frac12 \int_0^1 t^{k-1}(1-t)^{k-1}\,dt,$$ $$\frac{1}{k+1}+\cdots+\frac{1}{2k} = \int_0^1 \frac{1-x^{2k}}{1+x}\, dx$$ and $$\sum_{k=1}^\infty \frac{1}{(-2)^k}t^{k-1}(1-t)^{k-1}(1-x^{2k}) = -\frac{1}{2+t(1-t)}+\frac{x^2}{2+t(1-t)x^2}.$$ Combining all these together, we get $$\sum_{k=1}^\infty\frac1{k(-2)^k\binom{2k}k}\left(\frac1{k+1}+\ldots+\frac1{2k}\right) = I_1 + I_2,$$ where $$I_1 := -\int_0^1\frac{dx}{2(1+x)}\int_0^1 dt\,\frac{1}{2+t(1-t)} = -\frac13 \log(2)^2,$$ $$I_2 := \int_0^1 dx \frac{x^2}{2(1+x)} \int_0^1 dt\, \frac{1}{2+t(1-t)x^2} = \frac23\log(2)^2 - \frac{\pi^2}{36}.$$ So, $$I_1 + I_2 = \frac13\log(2)^2 - \frac{\pi^2}{36}.$$
• Could you kindly illustrate how to simplify the integral $I_2$？ Thank you！ Commented Feb 18, 2021 at 15:14