Infinite sum of $\sum_{n=1}^\infty \prod_{k=1}^n \frac{2k+1}{4k}$

Question

Here I have this sum, $\displaystyle \sum_{n=1}^\infty \displaystyle \prod_{k=1}^n \dfrac{2k+1}{4k}$.

I have no idea how to sum this up. Any help would be appreciated!

Answer 1

Is this true: $$\prod_{k=1}^n\frac{2k+1}{4k}=\frac{\prod_{k=1}^n2k+1}{\prod_{k=1}^n4k}=\frac{\prod_{k=1}^n2k+1}{4^n\prod_{k=1}^nk}=\frac{\Gamma(2n+1)}{4^n\Gamma(n)}$$

EDIT as pointed out by another user: $$\frac{\prod_{k=1}^n2k+1}{4^n\prod_{k=1}^nk}=\frac{(2n+1)!!}{4^nn!}$$