21
$\begingroup$

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:

$\endgroup$
  • 5
    $\begingroup$ 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 Jan 30 '18 at 12:28
  • 7
    $\begingroup$ known or not, Mathematica immediately evaluates it: link to Wolfram Alpha $\endgroup$ – Carlo Beenakker Jan 30 '18 at 12:49
  • 2
    $\begingroup$ Can it be interpreted as an expected value? $\endgroup$ – Michael Hardy Jan 31 '18 at 0:17
47
$\begingroup$

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 $a=1/2, b=0$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 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 Jan 31 '18 at 3:17
  • 1
    $\begingroup$ 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 Jan 31 '18 at 4:04
5
$\begingroup$

Use $\binom{n+k}{k}\binom{n}k$ in the sum. Define the functions $$F(n,k)=(-1)^k\frac{2n+1}{2k+1}\binom{n+k}k\binom{n}{k}, \qquad G(n,k)=\frac{(-1)^{k-1}}{n+1}\binom{n+k}{k-1}\binom{n}{k-1}.$$ 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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.