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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:

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    $\begingroup$ It may appear in a different form. E.g., notice that $\binom{n+k}{n-k}\binom{2k}{k}=\binom{n+k}{n}\binom{n}{k}$. $\endgroup$
    – Max Alekseyev
    Commented Jan 30, 2018 at 12:28
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    $\begingroup$ known or not, Mathematica immediately evaluates it: link to Wolfram Alpha $\endgroup$
    – Carlo Beenakker
    Commented Jan 30, 2018 at 12:49
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    $\begingroup$ Can it be interpreted as an expected value? $\endgroup$
    – Michael Hardy
    Commented Jan 31, 2018 at 0:17

2 Answers 2

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In terms of hypergeometric series, the sum is $_3F_2(-n, 1+n, 1/2;1,3/2;1)$ and the identity is a special case of Saalschütz's theorem (also called the Pfaff-Saalschütz theorem), one of the standard hypergeometric series identities.

A more general identity, also a special case of Saalschütz's theorem, is $$\sum_{k=0}^n (-1)^k\frac{a}{a+k}\binom{n+k+b}{n-k}\binom{2k+b}{k} = \binom{n+b-a}{n}\biggm/\binom{n+a}{n}.$$ The O.P.'s identity is the case $a=1/2, b=0$.

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    $\begingroup$ Thanks! I obtained the identity from the Clausen’s identity for the Legendre polynomials. A generalization to the associated Legendre functions produces $$\sum\limits_{k=m}^n\frac{(-1)^{k-m}}{2k+1}\binom{n+k}{n-k}\binom{2k}{k-m}=\frac{1}{2n+1}.$$ $\endgroup$
    – Zurab Silagadze
    Commented Jan 31, 2018 at 3:17
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    $\begingroup$ This identity also follows from the Saalschütz theorem (not immediately, but after some algebra) for the case $a=m+1/2$, $b=m+n+1$, $c=m+3/2$, because the sum now is $$\frac{\binom{n+m}{n-m}}{2m+1} {_3F_2}(m+1/2,m+n+1,-(n-m);m+3/2,2m+1;1).$$ $\endgroup$
    – Zurab Silagadze
    Commented Jan 31, 2018 at 4:04
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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.

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