Prove that $\sum_{k=0}^{n}\binom{n}{k}\frac{1}{2n-2k+1}=\frac{(2n)!!}{(2n+1)!!}$

Question

Prove that

\begin{equation} \sum_{k=0}^{n}\binom{n}{k}\frac{1}{2n-2k+1}=\frac{(2n)!!}{(2n+1)!!} \end{equation}

This sum appears in the orthogonalization of Legendre polynomials.

Answer 1

$$\begin{split} \sum_{k=0}^{n}\binom{n}{k}\frac{1}{2n-2k+1}&= \sum_{k=0}^{n}\binom{n}{k}\int_0^1 x^{2(n-k)}dx\\ &= \int_0^1\sum_{k=0}^{n}\binom{n}{k}x^{2(n-k)}dx\\ &= \int_0^1 \left(1+x^2\right)^ndx \end{split}$$

Answer 2

You could write it as a hypergeometric expression: $$ \frac{{}_{2}^{}{{{F_{1}^{}}}}\! \left(-n ,-n -\frac{1}{2};-n +\frac{1}{2};-1\right) }{2 n +1}$$