# Thread: Series : binomial coefficients

1. ## Series : binomial coefficients

Evaluate $\displaystyle \sum_{n=1}^\infty {2n \choose n}^{-1}$.

2. Spoiler:

For $\displaystyle n\geq 1$, we have $\displaystyle \frac{1}{{2n\choose n}}=\frac{n!n!}{(2n)!}=\frac{\Gamma(n+1)\Gamma(n+1 )}{\Gamma(2n+1)}$ $\displaystyle =\frac{n^2\Gamma(n)}{2n\Gamma(2n)}=\frac{n}{2}B(n, n)$ where $\displaystyle B$ is Euler's Beta function, hence $\displaystyle \frac{1}{{2n\choose n}}=\frac{n}{2}\int_0^1 x^{n-1}(1-x)^{n-1}dx$.

Since everything is positive, we may interchange sum and integral:

$\displaystyle \sum_{n=1}^\infty \frac{1}{{2n\choose n}}=\frac{1}{2}\int_0^1\left(\sum_{n=1}^\infty n(x(1-x))^{n-1}\right)dx$

hence

$\displaystyle \sum_{n=1}^\infty \frac{1}{{2n\choose n}} = \frac{1}{2}\int_0^1\frac{dx}{(1-x+x^2)^2}$.

Now, we can apply usual methods; Maple gives the value $\displaystyle \frac{1}{3}+\frac{2}{9\sqrt{3}}\pi$.

3. Great! My solution as well.

4. Originally Posted by Bruno J.
Great! My solution as well.
I'd be interested to see if there is a method by inverting the series expansion for arcsine.