How to evaluate $1+\frac{2^2}{3!}+\frac{3^2}{5!}+\frac{4^2}{7!}+\cdots$? I learnt that $\displaystyle \sum_{n=0}^{\infty} \frac{n+1}{(2n+1)!} = \frac{e}{2}$.
I am wondering what the closed form for $\displaystyle \sum_{n=0}^{\infty} \frac{(n+1)^2}{(2n+1)!}$ is. 
I tried using the fact that $ 1+3+5+\cdots+(2n-1) = n^2$, but it was not fruitful.
Could people give me some hints on how to approach this problem?
To view the general formula, please visit The value of $\sum_{n=0}^{\infty}\frac{(n+1)^k}{(2n+1)!}$, where $k\in\mathbb{W}$

$$\sinh x=\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)!}$$
differentiating and multiplying by $x$:
$$x\cosh x=\sum_{n=0}^\infty\frac{(2n+1)x^{2n+1}}{(2n+1)!}.$$
Again:
$$x^2\sinh x+x\cosh x=\sum_{n=0}^\infty\frac{(2n+1)^2x^{2n+1}}{(2n+1)!}.$$
Find $a$, $b$ and $c$ such that
$$a+b(2n+1)+c(2n+1)^2=(n+1)^2.$$
Then
$$a\sinh x+bx\cosh x+c(x^2\sinh x+x\cosh x)
=\sum_{n=0}^\infty\frac{(n+1)^2x^{2n+1}}{(2n+1)!}.$$
Now set $x=1$.
