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?
 A: I don't know if it's a bad manner to post the solution of own question but anyway thanks to Brian M. scott for his comment and using his suggestion: $\frac1{k+1}\binom{n}k=\frac1{n+1}\binom{n+1}{k+1}$ \begin{align}
\sum_{k=0}^n\frac{1}{k+1}{\binom n k}^2&=\sum_{k=0}^n\frac{1}{n+1}\binom{n+1}{k+1}\binom{n}{k}\\
&=\frac{1}{n+1}\sum_{k=0}^n\binom{n+1}{n-k}\binom{n}{k}\\
&=\frac{1}{n+1}\times\frac{(2n+1)!}{n!(n+1)!}\\
&=\frac{(2n+1)!}{{(n+1)!}^2}
\end{align}
A: Binomial theorem:
$$(1+t)^n=\sum_{k=0}^{n}{n \choose k} t^k~~~~(1)$$
Integrate w. r. t. $t$ from $t=0$ to $t=x$,
$$\frac{(1+x)^{n+1}-1}{n+1}=\sum_{k=0}^{n} \frac{n \choose k}{1+k}x^{1+k}~~~~(2)$$
Let $t=1/x$ in (1), to get
$$x^{-n}(1+x)^n=\sum_{k=0}^{n}{ n \choose k} x^{-k}~~~~(3)$$
Multiply (2) and (3) and retain only the terms of $x$
$$x \sum_{k=0}^{n}\frac{{n\choose k}^2}{1+k}+...+...+...=\frac{x^{-n}(1+x)^{2n+1}-x^{-n}(1+x)^n}{n+1}.$$
$$\implies  \sum_{k=0}^{n}\frac{{n\choose k}^2}{1+k}=[x^{n+1}] \frac{(1+x)^{2n+1}-(1+x)^n}{n+1}=\frac{2n+1\choose n+1}{n+1}=\frac{{2n+1\choose n}}{n+1}$$
Note that $[x^j] f(x)$ means coefficient of $x^j$ in $f(x)$.
A: 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 $\frac{1}{n+1}\binom{2n+1}{n+1}=\frac{(2n+1)!}{(n+1)!(2n+1-n-1)!}\frac{1}{n+1}=\frac{(2n+1)!}{{\left((n+1)!\right)}^2}$
A: As a consequence of the binomial theorem we have that $$\sum_{k=0}^{n}\dbinom{n}{k}\frac{x^{k+1}}{k+1}=\frac{\left(1+x\right)^{n+1}-1}{n+1}$$ hence, by the integral representation of the binomial coefficient, we have $$\sum_{k=0}^{n}\dbinom{n}{k}^{2}\frac{1}{k+1}=\frac{1}{2\pi i}\oint_{\left|z\right|=\varepsilon}\left(1+z\right)^{n}\sum_{k=0}^{n}\dbinom{n}{k}\frac{z^{-k-1}}{k+1}dz$$ $$=\frac{1}{n+1}\left[\frac{1}{2\pi i}\oint_{\left|z\right|=\varepsilon}\frac{\left(1+z\right)^{2n+1}}{z^{n+1}}dz-\frac{1}{2\pi i}\oint_{\left|z\right|=\varepsilon}\left(1+z\right)^{n}dz\right]=\color{red}{\frac{\dbinom{2n+1}{n}}{n+1}}.$$
