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 !!
 A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

The Question:
  $\ds{\sum_{k = 0}^{\infty}{n + k \choose 2k }{2k \choose k}
     {\pars{-1}^{k} \over k + 1} =\, ?}$

In using the combinatoric number definitions we note that
$$
{n + k \choose 2k }{2k \choose k}{1 \over k + 1} =
{1 \over n}{n \choose k}{n + k \choose n - 1}
$$
such that
\begin{align}
&\color{#f00}{\sum_{k = 0}^{\infty}
{n + k \choose 2k }{2k \choose k}{\pars{-1}^{k} \over k + 1}} =
{1 \over n}\sum_{k = 0}^{n}{n \choose k}{n + k \choose n - 1}\pars{-1}^{k}
\\[3mm] = &\
{1 \over n}\sum_{k = 0}^{n}{n \choose k}\pars{-1}^{k}\ \overbrace{%
\oint_{\verts{z} = 1}{\pars{1 + z}^{n + k} \over z^{n}}\,
{\dd z \over 2\pi\ic}}^{\ds{{n + k \choose n - 1}}}\ =\
{1 \over n}\oint_{\verts{z} = 1}{\pars{1 + z}^{n} \over z^{n}}\
\overbrace{\sum_{k = 0}^{n}{n \choose k}\pars{-1 - z}^{k}}^{\ds{\pars{-z}^{n}}}\
\,{\dd z \over 2\pi\ic}
\\[3mm] & =
{\pars{-1}^{n} \over n}
\oint_{\verts{z} = 1}\pars{1 + z}^{n}\,{\dd z \over 2\pi\ic} =
\color{#f00}{0}
\end{align}
A: 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.
