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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.

Your comments are welcome!

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  • $\begingroup$ 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 $$\sum_{k=1}^\infty\frac1{k^32^k\binom{3k}k}=\pi G+\frac{\log^22}6-\frac{\pi^2}{24}\log2-\frac{33}{16}\zeta(3).$$ $\endgroup$ Commented Feb 14, 2021 at 22:49
  • $\begingroup$ 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 $$\sum_{k=1}^\infty\frac1{k^3(-2)^k\binom{2k}k}=\frac{\log^32}6-\frac{\zeta(3)}4.$$ $\endgroup$ Commented Feb 14, 2021 at 23:07

1 Answer 1

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Here is a proof of the first identity. The others can probably be done similarly.

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, $$I_1 + I_2 = \frac13\log(2)^2 - \frac{\pi^2}{36}.$$

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  • $\begingroup$ Could you kindly illustrate how to simplify the integral $I_2$? Thank you! $\endgroup$
    – C. WANG
    Commented Feb 18, 2021 at 15:14
  • $\begingroup$ @C.WANG: I did not bother to do this manually as Maple computes this double integral routinely. The answer it gives is a bit longer, but it can be reduced to the stated form using the properties of dilogarithm and arctanh functions. $\endgroup$ Commented Feb 18, 2021 at 17:37

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