For each $n=0,1,2,\ldots$, the harmonic number $H_n$ is given by $$H_n:=\sum_{0<k\le n}\frac1k.$$

In 2016 I conjectured that

$$\sum_{k=1}^\infty\frac{3H_{k-1}^2+4H_{k-1}/k}{k^2\binom{2k}k}=\frac{\pi^4}{360}.\tag{1}$$ It is easy to check this numerically since the series converges rapidly.

Actually, $(1)$ was motivated by my following conjectural congruences $$p\sum_{k=1}^{p-1}\frac{3H_{k-1}^2+4H_{k-1}/k}{k^2\binom{2k}k} \equiv -3\frac{H_{p-1}}{p^2} - \frac{p^2}5B_{p-5}\pmod{p^3}\tag{2}$$


$$\sum_{k=1}^{p-1}\left(3H_k^2-4\frac{H_k}k\right)\frac{\binom{2k}k}k \equiv 6\frac{H_{p-1}}{p^2}+\frac 85p^2B_{p-5}\pmod{p^3},\tag{3}$$ where $p$ is any prime greater than $3$, and $B_0,B_1,\ldots$ are the Bernoulli numbers. It is well known that $$H_{p-1}\equiv-\frac{p^2}3B_{p-3}\pmod{p^3}\ \ \ \text{for any prime}\ p > 3.$$

I have no idea how to prove my conjectural identity $(1)$. I have inquired several mathematicians concerning $(1)$ but got no solution.

Question. How to prove the conjectural identity $(1)$?

Your comments towards its solution are welcome!

  • $\begingroup$ The questions arise: what for? where is that identity applied? BTW, geometry means "the earth measurement". $\endgroup$ – user64494 May 27 '19 at 15:00
  • $\begingroup$ $\pi^4$ comes up in the computation of the volume or surface area of the sphere in $\mathbb R^8$. $\endgroup$ – Ryan Budney May 27 '19 at 15:03

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