TITLE: Prove that $\sum_{j=0}^{n}H_j{n\choose j}^2={2n\choose n}\left(2H_n-H_{2n}\right)$
QUESTION [8 upvotes]: 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.

REPLY [13 votes]: 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 $$\sum_{j=0}^{n}\dbinom{n}{j}^{2}H_{j}=\dbinom{2n}{n}\left(2H_{n}-H_{2n}\right).$$