Show that $\sum_{n=1}^\infty\frac{n}{(2n+1)!!}={1\over2}$

Question

Problem. Show that $$\sum_{n=1}^\infty\frac{n}{3\cdot5\cdot\cdots\cdot(2n+1)}={1\over2}$$

The given series can be re-expressed as: $$\sum_{n=1}^\infty\frac{n}{3\cdot5\cdot\cdots\cdot(2n+1)}=\sum_{n=1}^\infty\frac{2^nn!n}{(2n+1)!}$$

I first attempted with binomial series but it was difficult to change the form of binomial. Could you please give me some ideas? Thanks.

Answer 1

Hint:$$\frac n{(2n+1)!!}=\frac12\left(\frac1{(2n-1)!!}-\frac1{(2n+1)!!}\right).$$

Answer 2

Use $$\frac{n}{(2n+1)!!}=\frac {1}{2}\left(\frac{1}{(2n-1)!!}-\frac{1}{(2n+1)!!}\right)$$ and the telescopic summation.

The needed sum is equal to $$\frac{1}{2}\lim_{n\rightarrow+\infty}\left(1-\frac{1}{(2n+1)!!}\right)=\frac{1}{2}.$$