TITLE: New binomial coefficient identity?
QUESTION [21 upvotes]: Is the following identity known?
$$\sum\limits_{k=0}^n\frac{(-1)^k}{2k+1}\binom{n+k}{n-k}\binom{2k}{k}=
\frac{1}{2n+1}$$
I have not found it in the following book:

Henry Wadsworth Gould, Combinatorial identities: a standardized set of tables listing 500 binomial coefficient summations, Morgantown, West Virginia, 1972.

REPLY [7 votes]: Use $\binom{n+k}{k}\binom{n}k$ in the sum. Define the functions
$$F(n,k)=(-1)^k\frac{2n+1}{2k+1}\binom{n+k}k\binom{n}{k}, \qquad
G(n,k)=2\cdot(-1)^{k-1}\binom{n+k}{k-1}\binom{n}{k-1}.$$
Then $F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k)$. Sum over all integers $k$ to obtain $$f(n+1)-f(n)=0$$ where $f(n)=\sum_kF(n,k)$ is your sum. Since $f(0)=1$, the identity follows.
This method is called the Wilf-Zeilberger technique of summation routine.