How to prove $\sum_{s=0}^{m}{2s\choose s}{s\choose m-s}\frac{(-1)^s}{s+1}=(-1)^m$? Question: How to prove the following identity?
$$
\sum_{s=0}^{m}{2s\choose s}{s\choose m-s}\frac{(-1)^s}{s+1}=(-1)^m.
$$
I'm also looking for the generalization of this identity like 
$$
\sum_{s=k}^{m}{2s\choose s}{s\choose m-s}\frac{(-1)^s}{s+1}=?
$$
Proofs, hints, or references are all welcome.

Suppose we seek to verify that
$$\sum_{s=0}^m {2s\choose s} {s\choose m-s} 
\frac{(-1)^s}{s+1} = (-1)^m$$
without using the generating function of the Catalan numbers.
Re-write the sum as follows:
$$\sum_{s=0}^m {2s\choose m} {m\choose s} 
\frac{(-1)^s}{s+1}$$
which is
$$\frac{1}{m+1}
\sum_{s=0}^m {2s\choose m} {m+1\choose s+1} (-1)^s$$
which turns into
$$- \frac{1}{m+1}
\sum_{s=1}^{m+1} {2s-2\choose m} {m+1\choose s} (-1)^{s}
\\ = \frac{1}{m+1} {-2\choose m}
- \frac{1}{m+1}
\sum_{s=0}^{m+1} {2s-2\choose m} {m+1\choose s} (-1)^{s}
\\ = \frac{1}{m+1} (-1)^m \frac{(m+1)!}{m!}
- \frac{1}{m+1}
\sum_{s=0}^{m+1} {2s-2\choose m} {m+1\choose s} (-1)^{s}
\\ = (-1)^m
- \frac{1}{m+1}
\sum_{s=0}^{m+1} {2s-2\choose m} {m+1\choose s} (-1)^{s}.$$
Now introduce
$${2s-2\choose m} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+1}} (1+z)^{2s-2} \; dz.$$
We get for the sum
$$- \frac{1}{m+1} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+1}} \frac{1}{(1+z)^2}
\sum_{s=0}^{m+1} {m+1\choose s} (-1)^{s} (1+z)^{2s}  \; dz
\\ = -\frac{1}{m+1} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+1}} \frac{1}{(1+z)^2}
(1-(1+z)^2)^{m+1}\; dz
\\ = \frac{(-1)^m}{m+1} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+1}} \frac{1}{(1+z)^2}
(z^2+2z)^{m+1} \; dz
\\ = \frac{(-1)^m}{m+1} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{(1+z)^2}
(z+2)^{m+1} \; dz
= 0.$$
This concludes the argument.
