i was messing around whith some trig identities and i came across this equation:
$$\sum_{k=0}^n\frac{1}{(2k)!(2(n-k)+1)!}=\frac{1}{2}\frac{2^{2n+1}}{(2n+1)!}\quad\quad(1)$$ This formula becomes pretty when choosing a value for n. Let n be 2, it gives: $$\frac{1}{0!\;5!}+\frac{1}{1!\;4!}+\frac{1}{2!\;3!}=\frac{1}{2}\frac{2^5}{5!}$$ The reason this happens is due to the trigonometric identity $\cos(x)\sin(x)=\frac{1}{2}\sin(2x)$, expanding in Taylor series both sides: $$\sum_{i=0}^\infty\left(\frac{(-1)^{i}}{(2i)!}x^{2i}\right)\sum_{j=0}^\infty\left(\frac{(-1)^{j}}{(2j+1)!}x^{2j+1}\right) = \frac{1}{2}\sum_{n=0}^\infty\left(\frac{(-1)^{n}}{(2n+1)!}(2x)^{2n+1}\right)$$ expandind the left hand side and grouping by the powers of x leads to equation (1).
Is it possible to find another proof of this identity (maybe without reliying on trig id.)?