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 $x=1$.
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.
Hint:
Let $(n+1)^2=A+B(2n+1)+C(2n+1)(2n)$
so that $$\dfrac{(n+1)^2}{(2n+1)!}=\dfrac A{(2n+1)!}+\dfrac B{(2n)!}+\dfrac C{(2n-1)!}$$
$$n^2+2n+1=4Cn^2+n(2C+2B)+A+B$$
Comparing the constants, $4C=1\iff C=?$
Comparing the coefficients of $n,$ $$2=2C+2B\iff B=?$$
Compare the coefficients of $n^2$ to find $A$
Finally use are the values of $e^x+e^{-x},e^x-e^{-x}?$
$$\sum_{n=0, \text{odd}}^{\infty} \frac{(\frac{n+1}{2})^2}{n!}$$
$$=\frac{1}{4} \sum_{n=0, \text{odd}}^{\infty} \frac{n^2+2n+1}{n!}$$
$$=\frac{1}{4} \left( \sum_{n=0,\text{odd}}^{\infty} \frac{n}{(n-1)!}+2\sum_{n=0,\text{odd}}^{\infty} \frac{1}{(n-1)!}+\sum_{n=0,\text{odd}}^{\infty} \frac{1}{n!} \right)$$
Now note:
$$\sum_{n=0, \text{odd}}^{\infty} \frac{1}{n!}$$
$$=\sum_{n=0}^{\infty} \frac{1}{2} \frac{1^n-(-1)^n}{n!} $$
$$=\frac{e^{1}-e^{-1}}{2}= \sinh 1$$
Also note:
$$S :=\sum_{n=0, \text{odd}}^{\infty} \frac{1}{(n-1)!}$$
$$=\sum_{n=0,\text{even}}^{\infty} \frac{1}{n!}$$
$$=\sum_{n=0}^{\infty} \frac{1}{2} \frac{1+(-1)^n}{n!}$$
$$=\cosh 1$$
Finally,
$$\sum_{n=0, \text{odd}}^{\infty} \frac{n}{(n-1)!}$$
$$=\sum_{n=0,\text{odd}}^{\infty} \frac{n-1}{(n-1)!}+\sum_{n=0,\text{odd}}^{\infty} \frac{1}{(n-1)!}$$
$$=S+\sum_{n=0,\text{odd}}^{\infty} \frac{(n-1)}{(n-1)!}$$
$$=S+\sum_{n=3,\text{odd}}^{\infty} \frac{(n-1)}{(n-1)!}$$
$$=S+\sum_{n=3,\text{odd}}^{\infty} \frac{1}{(n-2)!}$$
$$=S+\sum_{n=1,\text{odd}}^{\infty} \frac{1}{n!}$$
$$=S+\sum_{n=0,\text{odd}}^{\infty} \frac{1}{n!}$$
$$=S+\sinh 1$$
$$=\cosh 1+\sinh 1$$
This gives an answer of,
$$\frac{1}{4} \left(3\cosh 1+2\sinh 1 \right)$$
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{n = 0}^{\infty}{\pars{n + 1}^{2} \over \pars{2n + 1}!} & = {1 \over 4}\sum_{n = 0}^{\infty}{\bracks{\pars{2n + 1} + 1}^{\, 2} \over \pars{2n + 1}!} = {1 \over 4}\sum_{n = 0}^{\infty}{\pars{n + 1}^{\, 2} \over n!}\,{1 - \pars{-1}^{n} \over 2} \\[5mm] & = {1 \over 8}\left.\sum_{n = 0}^{\infty}{n^{2} + 2n + 1 \over n!} \,x^{n}\,\right\vert_{\ x\ =\ -1}^{\ x\ =\ \phantom{-}1} \\[5mm] & = \left.{1 \over 8}\bracks{\pars{x\,\partiald{}{x}}^{2} + 2x\,\partiald{}{x} + 1} \sum_{n = 0}^{\infty}{x^{n}\over n!} \,\right\vert_{\ x\ =\ -1}^{\ x\ =\ \phantom{-}1} \\[5mm] & = \left.{1 \over 8}\bracks{\pars{x\,\partiald{}{x}}^{2} + 2x\,\partiald{}{x} + 1}\expo{x} \,\right\vert_{\ x\ =\ -1}^{\ x\ =\ \phantom{-}1} \\[5mm] & = \left.{1 \over 8}\pars{x^{2} + 3x + 1}\expo{x} \,\right\vert_{\ x\ =\ -1}^{\ x\ =\ \phantom{-}1} \\[5mm] & = {1 \over 8}\,5\expo{} - {1 \over 8}\,\pars{-\expo{-1}} = \bbx{5\expo{2} + 1 \over 8\expo{}} \approx 1.7449 \end{align}
