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 $- 1 - n<0$

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:
$$\frac{1}{n+1}  [z^n] (1+z)^{n+1}
\sum_{k\ge 0} {n+k\choose k} z^k (-1)^k
\\ = \frac{1}{n+1}  [z^n] (1+z)^{n+1}
\frac{1}{(1+z)^{n+1}}
= \frac{1}{n+1}  [z^n] 1 = [[n=0]].$$
