Exercise from Comtet's Advanced Combinatorics: prove $27\sum_{n=1}^{\infty }1/\binom{2n}{n}=9+2\pi \sqrt{3}$

Solution 1:

Another one, which has been mentioned multiple times on this site (search for Wallis' product) is the following:

$$\displaystyle \int_{0}^{\pi/2} \sin^{2n-1} x \ dx = \dfrac{2 \cdot 4 \cdot 6 \cdots (2n-2)}{1 \cdot 3 \cdot 5 \cdots (2n-1)} \ \ n \gt 1$$

Now $$\displaystyle \dfrac{2 \cdot 4 \cdot 6 \cdots (2n-2)}{1 \cdot 3 \cdot 5 \cdots (2n-1)} = \dfrac{2^{2n}}{2n{2n \choose n}} = \dfrac{2^{2n-1}}{n{2n \choose n}}$$

Thus

$$\displaystyle \int_{0}^{\pi/2} n\left(\dfrac{\sin x}{2}\right)^{2n-1} \ dx = \dfrac{1}{{2n \choose n}}$$

Now the sum can be found easily.

The formula by Alfred Van der Pooten can be proved using this approach too.

A proof of Wallis Product is given here: http://crypto.stanford.edu/pbc/notes/pi/wallis.xhtml

For a (closely related) problem on this site using this, see here: Summing the series $(-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k+1}$

Solution 2:

$1/\binom{2n}{n} = (2n+1) \int_0^1 x^n(1-x)^n dx$. Sum these over $n$ to get the integral of a rational function.