Value of this convergent series: $\frac{1}{3!}+\frac2{5!}+\frac3{7!}+\frac{4}{9!}+\cdots$

Question

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!

Answer 1

One may write $$ \begin{align} \sum_{n=0}^\infty\frac{n}{(2n+1)!}&=\frac12\sum_{n=0}^\infty\frac{(2n+1)-1}{(2n+1)!} \\\\&=\frac12\sum_{n=0}^\infty\frac{1}{(2n)!}-\frac12\sum_{n=0}^\infty\frac{1}{(2n+1)!} \\\\&=\frac12\left(\frac{e+e^{-1}}2 \right)-\frac12\left(\frac{e-e^{-1}}2 \right) \\\\&=\frac12\cdot e^{-1} \end{align} $$

Answer 2

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