Calculate $\sum_{0 \le k } \binom{n+k}{2k} \binom{2k}{k} \frac{(-1)^k}{k+1}$ 
Calculate $$\sum_{0 \le k } \binom{n+k}{2k} \binom{2k}{k}
 \frac{(-1)^k}{k+1}$$

My approach
$$\sum_{0 \le k } \binom{n+k}{2k} \binom{2k}{k}
 \frac{(-1)^k}{k+1} = \\
\sum_{0 \le k } \binom{n+k}{k} \binom{n}{k}
 \frac{(-1)^k}{k+1} = \\
\frac{1}{n+1}\sum_{0 \le k } \binom{n+k}{k} \binom{n+1}{k+1}(-1)^k = \\
\frac{1}{n+1}\sum_{0 \le k } \binom{k - 1 - n - k}{k} \binom{n+1}{k+1}
$$
But unfortunately I have stucked, I don't know how I can finish that...
The main obstacle which I see is
$$\binom{- 1 - n}{k} $$
is looks so dangerous because $- 1 - n<0$
 A: Starting from
$$\sum_{k=0}^n {n+k\choose 2k} {2k\choose k} \frac{(-1)^k}{k+1}$$
for a self-contained answer we observe that
$${n+k\choose 2k} {2k\choose k} =
\frac{(n+k)!}{(n-k)! \times k! \times k!}
= {n+k\choose k} {n\choose k}$$
so we find
$$\sum_{k=0}^n {n+k\choose k} {n\choose k} \frac{(-1)^k}{k+1}$$
which is
$$\frac{1}{n+1}
\sum_{k=0}^n {n+k\choose k} {n+1\choose k+1} (-1)^k
\\ = \frac{1}{n+1}
\sum_{k=0}^n {n+k\choose k} {n+1\choose n-k} (-1)^k
\\ = \frac{1}{n+1}  [z^n] (1+z)^{n+1}
\sum_{k=0}^n {n+k\choose k} z^k (-1)^k.$$
The coefficient  extractor controls the  range (with $k\gt n$  we will
always  have $[z^n]  (1+z)^{n+1}  z^k =  0$) and  we  may continue  by
extending $k$ to infinity:
$$\frac{1}{n+1}  [z^n] (1+z)^{n+1}
\sum_{k\ge 0} {n+k\choose k} z^k (-1)^k
\\ = \frac{1}{n+1}  [z^n] (1+z)^{n+1}
\frac{1}{(1+z)^{n+1}}
= \frac{1}{n+1}  [z^n] 1 = [[n=0]].$$
A: 
Your approach is  also fine.   Starting  with  the  last  line  we  obtain  for  $n\geq 0$:
\begin{align*}
\color{blue}{\frac{1}{n+1}}&\color{blue}{\sum_{k\geq 0}\binom{-n-1}{k}\binom{n+1}{k+1}}\tag{1}\\
&=\frac{1}{n+1}\sum_{k=0}^n\binom{-n-1}{k}\binom{n+1}{n-k}\tag{2}\\
&=\frac{1}{n+1}\binom{0}{n}\tag{3}\\
&\,\,\color{blue}{=[[n=0]]}\tag{4}
\end{align*}

Comment:

*

*In  (1)   we  note $\binom{-n-1}{k}$ is a  valid expression following the more general definition of binomial coefficients for $\alpha\in\mathbb{C},k\in\mathbb{Z}$:

\begin{align*}
\binom{\alpha}{k}=
\begin{cases}
\frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}&\qquad  k\geq  0\\
0&\qquad k<0
\end{cases}
\end{align*}
$\qquad$This  definition is given for instance as (5.1)   in  Concrete Mathematics  by R.L.  Graham, D.E.  Knuth and O. Patashnik.

*

*In (2) we use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$. We also set the upper limit of the sum to $n$, noting that $\binom{n+1}{n-k}=0$ if $k>n$.


*In (3) we apply Chu-Vandermonde's identity.


*In (4) we use Iverson brackets as compact notation.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\sum_{0\ \leq\ k}{n + k \choose 2k}{2k \choose k}{\pars{-1}^{k} \over k + 1}} =
\sum_{k\ =\ 0}^{\infty}{n + k \choose n - k}\bracks{{-1/2 \choose k}
\pars{-4}^{k}}\pars{-1}^{k}\int_{0}^{1}t^{k}\,\dd t
\\[5mm] = &\
\int_{0}^{1}\sum_{k\ =\ 0}^{\infty}{-1/2 \choose k}
\braces{\bracks{z^{n - k}}\pars{1 + z}^{n + k}}\pars{4t}^{k}\,\dd t
\\[5mm] = &\
\bracks{z^{n}}\pars{1 + z}^{n}
\int_{0}^{1}\sum_{k\ =\ 0}^{\infty}{-1/2 \choose k}
\bracks{4z\pars{1 + z}t}^{k}\,\dd t
\\[5mm] = &\
\bracks{z^{n}}\pars{1 + z}^{n}\
\underbrace{\int_{0}^{1}\bracks{1 + 4z\pars{1 + z}t}^{\, -1/2}\,\,\dd t}
_{\ds{1 \over 1 + z}}\ =\ \bracks{z^{n}}\pars{1 + z}^{n - 1} =
\bbx{\large \delta_{n0}}
\end{align}
