Where $H_0=0$ and $H_n$ is the n-th harmonic number.
Show that this sum has a closed form
$$\sum_{j=0}^{n}j{n\choose j}^2H_{n-j}={1\over 4}{2n\choose n}(4nH_{n}-2nH_{2n}-1)$$
I try:
$j{2n\choose n}={(2n)!\over (j-1)!(2n-j)!}$ look more messier!
I found two sum on Wikipedia site but it doesn't see more helpful at all!
$$\sum_{j=1}^{n}H_j=(n+1)(H_{n+1}-1)$$
$$\sum_{j=0}^{n}{n\choose j}^2=2^n$$
Any help? Thank you!
Using the Vandermonde Identity we have, $$\sum\limits_{j=0}^{n-1} \binom{n+x-1}{j}\binom{n}{n-j-1} = \binom{2n+x-1}{n-1}$$ i.e., after reindexing the summation and using the identity $\displaystyle \binom{n+x-1}{j-1} = \frac{j}{n+x}\binom{n+x}{j}$ we have,
$$\sum\limits_{j=0}^{n} j\binom{n+x}{j}\binom{n}{n-j} = (n+x)\binom{2n+x-1}{n-1}$$
Differentiating both sides w.r.t $x$ using $\displaystyle \frac{d}{dx} \binom{n+x}{j} = \binom{n+x}{j}(H_n(x) - H_{n-j}(x))$ where $\displaystyle H_n(x) = \sum\limits_{j=1}^{n} \frac{1}{j+x}$,
\begin{align*}&\sum\limits_{j=0}^{n} j\binom{n+x}{j}\binom{n}{n-j}(H_n(x) - H_{n-j}(x)) \\&= \binom{2n+x-1}{n-1}((n+x)(H_{2n-1}(x) - H_{n}(x)) + 1)\end{align*}
i.e., at $x = 0$ we have, \begin{align*} \sum\limits_{j=0}^{n} j\binom{n}{j}^2(H_n - H_{n-j}) &= \binom{2n-1}{n-1}(n(H_{2n-1} - H_{n}) + 1) \\ \implies \sum\limits_{j=0}^{n} j\binom{n}{j}^2H_{n-j} &= H_n\sum\limits_{j=0}^{n} j\binom{n}{j}^2 - \binom{2n-1}{n-1}(n(H_{2n-1} - H_{n}) + 1)\\&= \frac{1}{2}\binom{2n}{n}\left(2nH_n - nH_{2n} - \frac{1}{2}\right)\end{align*}
Alternative:
One may use the identity $\displaystyle \sum\limits_{j=1}^{k} \frac{(-1)^{j-1}}{j}\binom{n}{k-j} = \binom{n}{k}(H_n - H_{n-k}) \tag{*}$ to write:
\begin{align*} \sum\limits_{k=1}^{n} k\binom{n}{k}^2(H_n-H_{n-k}) &= \sum\limits_{k=1}^{n} \sum\limits_{j=1}^{k}\frac{(-1)^{j-1}k}{j}\binom{n}{k}\binom{n}{k-j} \tag{1}\\&= \sum\limits_{j=1}^{n}\sum\limits_{k=j}^{n}\frac{(-1)^{j-1}k}{j}\binom{n}{k}\binom{n}{k-j}\\&= n\sum\limits_{j=1}^{n}\frac{(-1)^{j-1}}{j}\sum\limits_{k=j}^{n}\binom{n-1}{k-1}\binom{n}{k-j}\\&= n\sum\limits_{j=1}^{n}\frac{(-1)^{j-1}}{j}\sum\limits_{k=0}^{n-j}\binom{n-1}{n-k-j}\binom{n}{k}\\&= n\sum\limits_{j=1}^{n}\frac{(-1)^{j-1}}{j}\binom{2n-1}{n-j}\\&= n\binom{2n-1}{n}(H_{2n-1} - H_{n-1}) \tag {2}\end{align*} where, in steps $(1)$ and $(2)$ we used identity $(*)$.
