A Conceptual Intepretation of the identity

Question

I came across the identity $$\sum_{k = 0}^{n}\frac{(-1)^{k}\binom{n}{k}}{2k+1} = \frac{2^{2n}(n!)^2}{(2n+1)!}$$ I tried it using the binomial theorem and integration and was able to prove it provided the identity $$∫^1_0(1−x^2)^ndx=\frac{(2n)!!}{(2n+1)!!}$$ $$$$MY TRY : $$(1-x^2)^n = \sum_{k = 0}^{n}(-1)^k\binom{n}{k}(x^2)^k $$ now integrating both sides gives us $$∫^1_0(1−x^2)^ndx = \sum_{k = 0}^{n}\frac{(-1)^{k}\binom{n}{k}}{2k+1} = \frac{(2n)!!}{(2n+1)!!} $$ A little manipulation gives us $$\frac{(2n)!!}{(2n+1)!!} = \frac{2^{2n}(n!)^2}{(2n+1)!}$$ All this seems much calculative, is there a conceptual way of explaining the identity i.e.. via probability or permutation and combination?

Answer 1

By the binomial transform proving such identity is equivalent to proving

$$ \sum_{k=0}^{n}(-4)^k \binom{n}{k}\frac{k!^2}{(2k+1)!} = \frac{1}{2n+1}$$

which looks scarier, but is really not due to

$$ \sum_{k=0}^{n}(-4)^k \binom{n}{k}B(k+1,k+1)=\int_{0}^{1}\sum_{k=0}^{n}\binom{n}{k}(4x(x-1))^k\,dx=\int_{0}^{1}(2x-1)^{2n}\,dx=\frac{1+(-1)^{2n}}{4n+2}. $$ Not really sure about a proper probabilistic interpretation, although.