2
$\begingroup$

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!

Improve this question
$\endgroup$
3
  • $\begingroup$ 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 $$\sum_{k=1}^\infty\frac1{k^32^k\binom{3k}k}=\pi G+\frac{\log^22}6-\frac{\pi^2}{24}\log2-\frac{33}{16}\zeta(3).$$ $\endgroup$
    – Zhi-Wei Sun
    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 $$\sum_{k=1}^\infty\frac1{k^3(-2)^k\binom{2k}k}=\frac{\log^32}6-\frac{\zeta(3)}4.$$ $\endgroup$
    – Zhi-Wei Sun
    Commented Feb 14, 2021 at 23:07
  • $\begingroup$ How do you prove the reciprocal cube Zete3 series? I have yet to see that one anywhere. $\endgroup$
    – CarP24
    Commented Nov 11 at 16:56

1 Answer 1

Reset to default
6
$\begingroup$

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

Improve this answer
$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Max Alekseyev
    Commented Feb 18, 2021 at 17:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .