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.
On
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.
On
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 $f(x):=\sum_{s=0}^{\infty}a_sx^s=\sum_{s=0}^\infty[y^s]f(y)x^s$
Let $$a_m=\sum_{s=0}^{m}\binom{2s}{s}\binom{s}{m-s}\dfrac{(-1)^s}{s+1}$$and let $P(x)=\displaystyle\sum_{m=0}^{\infty}a_m x^m$ be the generating function for $a_m$. We can rewrite $P(x)$ as $$\sum_{m=0}^{\infty}\sum_{s=0}^{m}\binom{2s}{s}\binom{s}{m-s}\frac{(-1)^s}{s+1}x^m=\sum_{s=0}^{\infty}\sum_{m=s}^{\infty}\binom{2s}{s}\binom{s}{m-s}\frac{(-1)^s}{s+1}x^m$$The second sum is really the same thing as a sum of $m-s$ from $0$ to $\infty$, so we can write it as so and take out the terms that don't involve $m$: $$\sum_{s=0}^{\infty}\binom{2s}{s}\frac{(-1)^s}{s+1}x^s\sum_{m-s=0}^{\infty}\binom{s}{m-s}x^{m-s}$$By the binomial theorem, the above is equal to $$\sum_{s=0}^{\infty}\binom{2s}{s}\frac{(-1)^s}{s+1}x^s (1+x)^s=\sum_{s=0}^{\infty}\binom{2s}{s}\frac{1}{s+1}(-x-x^2)^s$$From the generating function of the Catalan numbers, $\displaystyle\sum_{n=0}^{\infty}\binom{2n}{n}\dfrac{1}{n+1}x^n=\dfrac{1-\sqrt{1-4x}}{2x}$, this last expression is equal to $$\frac{1-\sqrt{1+4x+4x^2}}{-2x-2x^2}=\frac{-2x}{-2x-2x^2}=\frac{1}{x+1}=1-x+x^2-x^3+\cdots$$So to conclude, we know that $$\sum_{m=0}^{\infty}a_m x^m=\sum_{m=0}^{\infty}(-1)^m x^m\Leftrightarrow a_m=(-1)^m$$This method is known as the snake oil method. To evaluate the more general expression, you'll need to truncate some of the terms at the beginning of the sum.