Let $H_n$ the $n$th Harmonic numbers and $H_0=0.$
Prove that $$\sum_{j=0}^{n}H_j{n\choose j}^2={2n\choose n}\left(2H_n-H_{2n}\right)$$
I encounter this problem since 2012 and have verify numerically and not sure it is correct for sure. So can anybody help me to prove it.
Let $j\leq n $ and let us define $$H_{n}\left(x\right)=\sum_{k=1}^{n}\frac{1}{n+x}. $$ It is not difficult to prove that $$\frac{d}{dx}\dbinom{x+n}{j}=\dbinom{x+n}{j}\left(H_{n}\left(x\right)-H_{n-j}\left(x\right)\right) $$ so in particular $$\frac{d}{dx}\dbinom{x+n}{j}_{x=0}=\dbinom{n}{j}\left(H_{n}-H_{n-j}\right). $$ Now for the Chu-Vandermonde identity we have $$\sum_{j=0}^{n}\dbinom{n+x}{j}\dbinom{n}{n-j}=\dbinom{2n+x}{n} $$ so if we take the derivative we have $$\sum_{j=0}^{n}\dbinom{n+x}{j}\dbinom{n}{n-j}\left(H_{n}\left(x\right)-H_{n-j}\left(x\right)\right)=\dbinom{2n+x}{n}\left(H_{2n}\left(x\right)-H_{n}\left(x\right)\right) $$ then, if we take $x=0 $, $$H_{n}\sum_{j=0}^{n}\dbinom{n}{j}^{2}-\sum_{j=0}^{n}\dbinom{n}{j}^{2}H_{j}=\dbinom{2n}{n}\left(H_{2n}-H_{n}\right) $$ but since $$ \sum_{j=0}^{n}\dbinom{n}{j}^{2}=\dbinom{2n}{n} $$ we have $$\sum_{j=0}^{n}\dbinom{n}{j}^{2}H_{j}=\dbinom{2n}{n}\left(2H_{n}-H_{2n}\right).$$