Studying Representation Theory of finite groups I've bumped in the following identity:
$$\frac{n(n+1)}{2}=\sum_{i=1}^n\frac{(2i-1)!!(2n-2i+1)!!}{(2i-2)!!(2n-2i)!!}$$
My book suggests to prove it looking at the dimension of the representation of $S_n$ related to a particular partition of $\frac{n(n+1)}{2}$.
Now my question: someone has other ideas to prove the above equality? In particular I'm looking for an elementary proof (by combinatorics or by analytical instruments).
We have: $$\frac{(2i-1)!!}{(2i-2)!!}=\frac{(2i-1)!}{4^{i-1}(i-1)!^2}=\frac{2}{\pi}\int_{0}^{\pi/2}\sin^{2i}\theta\,d\theta=\frac{2}{\pi}\int_{0}^{1}\frac{x^{2i}}{\sqrt{1-x^2}}\,dx,$$ $$\frac{(2n-2i+1)!!}{(2n-2i)!!}=\frac{2}{\pi}(2n-2i+1)\int_{0}^{\pi/2}\sin^{2n-2i}\theta\,d\theta$$ hence: $$\frac{(2i-1)!!}{(2i-2)!!}\cdot\frac{(2n-2i+1)!!}{(2n-2i)!!}=\frac{2}{\pi\cdot i! (n-i)!}\Gamma\left(i+\frac{1}{2}\right)\Gamma\left(\frac{3}{2}+n-i\right)$$ and you identity follows from the generalized Chu-Vandermonde identity with $m=\frac{1}{2}$ and $n=\frac{3}{2}$.