<s>[INST] How to find the closed form of $\sum_{k=0}^n\frac{1}{k+1} {\binom n k}^2$ How to sum the following identity: $$\sum_{k=0}^n\frac{1}{k+1}{\binom n k}^2$$  I know the answer is $\frac{(2n+1)!}{{(n+1)!}^2}$ but I have tried a lot to find the closed form through any algebraic way or combinatorial but not able to find it out, any help?
 [/INST] consider this polynomial (or OGF).
$$
\sum_{k=0}^n \binom{n}{k}^2 \frac{1}{k+1}=[x^{n+1}]\sum_{k=0}^n \binom{n}{n-k}x^{n-k}\times \binom{n}{k}\frac{x^{k+1}}{k+1}
$$
It's the convolution(product) of $(1+x)^n$ and $F(x)$ where
$$
\begin{aligned}
F(x)&=\sum_{k=0}^n \binom{n}{k}\frac{x^{k+1}}{k+1}\\
F'(x)&=\sum_{k=0}^n \binom{n}{k}x^k=(1+x)^n\\
F(x)&=\frac{(1+x)^{n+1}}{n+1}
\end{aligned}
$$
Thus the polynomial is $(1+x)^n\frac{(1+x)^{n+1}}{n+1}$ and the $(n+1)$ -th coefficient is $\frac{1}{n+1}\binom{2n+1}{n+1}=\frac{(2n+1)!}{(n+1)!(2n+1-n-1)!}\frac{1}{n+1}=\frac{(2n+1)!}{{\left((n+1)!\right)}^2}$
</s>