# How to prove the identity $\sum_{k=1}^\infty\frac{3H_{k-1}^2+4H_{k-1}/k}{k^2\binom{2k}k}=\frac{\pi^4}{360}$?

For each $$n=0,1,2,\ldots$$, the harmonic number $$H_n$$ is given by $$H_n:=\sum_{0

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 $$p\sum_{k=1}^{p-1}\frac{3H_{k-1}^2+4H_{k-1}/k}{k^2\binom{2k}k} \equiv -3\frac{H_{p-1}}{p^2} - \frac{p^2}5B_{p-5}\pmod{p^3}\tag{2}$$

and

$$\sum_{k=1}^{p-1}\left(3H_k^2-4\frac{H_k}k\right)\frac{\binom{2k}k}k \equiv 6\frac{H_{p-1}}{p^2}+\frac 85p^2B_{p-5}\pmod{p^3},\tag{3}$$ 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 $$(1)$$?

• $\pi^4$ comes up in the computation of the volume or surface area of the sphere in $\mathbb R^8$. – Ryan Budney May 27 at 15:03