About a sum involving factorials.

Question

I would like to know if there is a closed form of $$\sum_{k=0}^{n}\frac{4^{k}}{\left(2k\right)!\left(n-k\right)!^{2}}.$$ Wolfram gives a strange closed form and, i.e., $$\frac{16^{n}\left(2n-\frac{1}{2}\right)!}{\sqrt{\pi}\left(2n!\right)^{2}}$$ and I'm not sure is right. If it's so, I have no idea how to prove it. Is it right? And how to prove this identity? Any help would be appreciate! Thank you.

Answer 1

It seems like you, too, have come across Jack's question from yesterday. Assuming the validty of his identity, your sum, which is the convolution (Cauchy product) formed when multiplying $\cos(1)$ and $\sum_{k\geq 0}\frac{(-1)^k}{k!^2 4^k}$ is equal to $$\frac{(4n)!}{(2n)!^{3}}$$