<s>[INST] 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 $k\in\mathbb{W}$
 [/INST] Note that 
$$\sinh x = \frac{e^x - e^{-x}}{2} = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots$$
Then
$$x \sinh x = x^2 + \frac{x^4}{3!} + \frac{x^6}{5!} + \cdots$$
Now we replace $x$ with $\sqrt{x}$ getting
$$\sqrt{x} \sinh \sqrt{x} = x + \frac{x^2}{3!} + \frac{x^3}{5!} + \cdots$$
We take the derivative
$$(\sqrt{x} \sinh \sqrt{x})' = 1 + \frac{2x}{3!} + \frac{3x^2}{5!} + \cdots$$
Then we multiply by $x$
$$x (\sqrt{x} \sinh \sqrt{x})' = x + \frac{2x^2}{3!} + \frac{3x^3}{5!} + \cdots$$
and take the derivative again
$$\left( x (\sqrt{x} \sinh \sqrt{x})' \right)' = 1 + \frac{2^2 x}{3!} + \frac{3^2 x^2}{5!} + \cdots$$
Finally, we set $x=1$
$$\left. \left( x (\sqrt{x} \sinh \sqrt{x})' \right)' \right|_{x=1} = 1 + \frac{2^2}{3!} + \frac{3^2}{5!} + \cdots$$
Thus, you only need to expand the left hand side analytically to find the value of the series.
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