# Series : binomial coefficients

• April 21st 2010, 12:07 PM
Bruno J.
Series : binomial coefficients
Evaluate $\sum_{n=1}^\infty {2n \choose n}^{-1}$.
• April 21st 2010, 03:01 PM
Laurent
Spoiler:

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

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

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

hence

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

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

• April 21st 2010, 05:11 PM
Bruno J.
Great! My solution as well. (Cool)
• April 21st 2010, 07:01 PM
Drexel28
Quote:

Originally Posted by Bruno J.
Great! My solution as well. (Cool)

I'd be interested to see if there is a method by inverting the series expansion for arcsine.