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

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

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}
$$