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.

This is just a supplement to  the nice answer of @tc2718. We show that it is convenient to use the  coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ of a series. We can write e.g.
$$\binom{n}{k}=[x^k](1+x)^n$$

We obtain
  \begin{align*}
 \sum_{s=0}^{m}&{2s\choose s}{s\choose m-s}\frac{(-1)^s}{s+1}\\
  &= \sum_{s=0}^{m}[u^{m-s}](1+u)^s(-1)^s[x^s]\frac{1-\sqrt{1-4x}}{2x}\tag{1}\\
  &= [u^{m}]\sum_{s=0}^{m}(1+u)^s(-u)^s[x^s]\frac{1-\sqrt{1-4x}}{2x}\tag{2}\\
  &= [u^{m}]\frac{1-\sqrt{1-4(1+u)(-u)}}{2(1+u)(-u)}\tag{3}\\
  &= [u^{m}]\frac{1-(1+2u)}{2(1+u)(-u)}\\
  &= [u^{m}]\frac{1}{1+u}\\
  &= [u^{m}]\sum_{s=0}^{\infty}(-u)^s\\
  &=(-1)^m
  \end{align*}

Comment:


*

*In (1) we use the coefficient of operator together with the series expansion of the Catalan-numbers:
$\frac{1-\sqrt{1-4x}}{2x}=\sum_{s=0}^{\infty}\frac{1}{s+1}\binom{2s}{s}x^s$

*In (2) we use the rule $[x^s]f(x)=[x^0]x^{-s}f(x)$

*In (3) we use the substitution rule $f(x):=\sum_{s=0}^{\infty}a_sx^s=\sum_{s=0}^\infty[y^s]f(y)x^s$
