TITLE: Value of this convergent series: $\frac{1}{3!}+\frac2{5!}+\frac3{7!}+\frac{4}{9!}+\cdots$
QUESTION [9 upvotes]: What is the value of- $$\frac{1}{3!}+\frac2{5!}+\frac3{7!}+\frac{4}{9!}+\cdots$$  I  wrote it as general term $\sum\frac{n}{(2n+1)!}$. As the series converges it should be telescopic (my thought). But i dont know how to proceed. I also know $\sum\frac{1}{n!}=e$ Any help /hints appreciated. Thanks!

REPLY [14 votes]: We know that
$$\frac{\sinh x}{x}=\sum_{n=0}^{\infty }\frac{x^{2n}}{(2n+1)!}$$
let $x\rightarrow \sqrt{x}$
$$\frac{\sinh \sqrt{x}}{\sqrt{x}}=\sum_{n=0}^\infty \frac{x^n}{(2n+1)!}$$
$$\left(\frac{\sinh \sqrt{x}}{\sqrt{x}}\right)'=\sum_{n=1}^\infty \frac{nx^{n-1}}
{(2n+1)!}$$
let $x=1$ to get what do you want