Summation of Binomial Coefficient: $\sum\binom{n+k}{2k} \binom{2k}k \frac{(-1)^l}{k+1}$ I am trying to solve this summation problem .
$$\sum\limits_{k = 0}^\infty  {\left( {\begin{array}{*{20}{l}}
{n + k}\\
{2k}
\end{array}} \right)} \left( {\begin{array}{*{20}{l}}
{2k}\\
k
\end{array}} \right)\frac{{{{( - 1)}^k}}}{{k + 1}}$$
It will be grateful if someone could help me !!

Suppose we seek to evaluate
$$\sum_{k=0}^n {n+k\choose 2k} {2k\choose k}\frac{(-1)^k}{k+1}
= \sum_{k=0}^n {n+k\choose n-k} {2k\choose k}\frac{(-1)^k}{k+1}.$$
Introduce
$${n+k\choose n-k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-k+1}} (1+z)^{n+k}\; dz.$$
This vanishes when $k\gt n$ and we  may extend the sum to infinity. We
obtain
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{n}
\sum_{k\ge 0} {2k\choose k}\frac{(-1)^k}{k+1} z^k (1+z)^k
\; dz.$$
We recognize the generating function of the Catalan numbers which is
$$\sum_{k\ge 0} {2k\choose k}\frac{1}{k+1} w^k
= \frac{1-\sqrt{1-4w}}{2w}.$$
This is certainly  analytic in $|w|<1/4.$ We  therefore fix $\epsilon$
so that $|z(1+z)|<1/4,$ for example take $\epsilon = 1/10.$
This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{n}
\frac{1-\sqrt{1+4z(1+z)}}{2\times (-1)\times z(1+z)}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+2}} (1+z)^{n-1}
\frac{\sqrt{(1+2z)^2}-1}{2}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+2}} (1+z)^{n-1}
\frac{2z}{2}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{n-1}
\; dz.$$
This is $$[z^n] (1+z)^{n-1} = 0.$$
Here we have used the branch of the logarithm with the cut on the negative real axis.
