Compute $\frac{1^2 t}{1!}+\frac{2^2 t^2}{3!}+\frac{3^2 t^3}{5!}+\frac{4^2 t^4}{7!}+\ldots+\frac{n^2 t^n}{(2n-1)!}+\ldots$ I have to compute $$\frac{1^2 t}{1!}+\frac{2^2 t^2}{3!}+\frac{3^2 t^3}{5!}+\frac{4^2 t^4}{7!}+\ldots+\frac{n^2 t^n}{(2n-1)!}+\ldots$$ I know that $\sinh t$ can be represented as a series1, but for that I require only odd powers $t^{2n-1}$, but I have no idea how to get them or what to do with the even part. Can anybody give me a hint how to start?
1 The expansion is $$\sinh x = x + \frac {x^3} {3!} + \frac {x^5} {5!} + \frac {x^7} {7!} +\cdots = \sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}.$$
 A: \begin{align*}
  \sqrt{t} \sinh \sqrt{t} &= \sum_{n=1}^{\infty} \frac{t^{n}}{(2n-1)!} \\
  (\sqrt{t} \sinh \sqrt{t})' &=\sum_{n=1}^{\infty} \frac{nt^{n-1}}{(2n-1)!} \\
  t(\sqrt{t} \sinh \sqrt{t})' &=\sum_{n=1}^{\infty} \frac{nt^{n}}{(2n-1)!} \\
  [t(\sqrt{t} \sinh \sqrt{t})']' &=
  \sum_{n=1}^{\infty} \frac{n^{2}t^{n-1}}{(2n-1)!} \\
  t[t(\sqrt{t} \sinh \sqrt{t})']' &=
  \sum_{n=1}^{\infty} \frac{n^{2}t^{n}}{(2n-1)!} \\
  \sum_{n=1}^{\infty} \frac{n^{2}t^{n}}{(2n-1)!} &=
  t\left[t
     \left(
       \frac{\sinh \sqrt{t}}{2\sqrt{t}}+
       \sqrt{t} \cosh \sqrt{t} \times \frac{1}{2\sqrt{t}}
     \right)
   \right]' \\
  &=t\left(
       \frac{\sqrt{t}}{2}\sinh \sqrt{t}+\frac{t}{2} \cosh \sqrt{t}
     \right)' \\
  &=t\left(
       \frac{1}{4\sqrt{t}}\sinh \sqrt{t}+
       \frac{\sqrt{t}}{2}\cosh \sqrt{t} \times \frac{1}{2\sqrt{t}}+
       \frac{1}{2} \cosh \sqrt{t}+
       \frac{t}{2} \sinh \sqrt{t} \times \frac{1}{2\sqrt{t}}
     \right) \\
  &= \frac{(t+1)\sqrt{t}}{4} \sinh \sqrt{t}+
     \frac{3t}{4} \cosh \sqrt{t}
\end{align*}
A: Let's carry this computation out for instructional purposes.
Let
$$f(t) = \sum_{n=1}^{\infty} \frac{t^n}{(2 n)!} = \cosh{\left (\sqrt{t}\right )}-1$$
Then the sum we seek is
$$2 \sum_{n=1}^{\infty} \frac{n^3 t^n}{(2 n)!} = \left (t \frac{d}{dt} \right )^3 f(t) = 2 t f'(t) + 6 t^2 f''(t) + 2 t^3 f'''(t)$$
$$f'(t) = \frac{\sinh{\left (\sqrt{t}\right )}}{2 \sqrt{t}} $$
$$f''(t) = -\frac14 t^{-3/2} \sinh{\left (\sqrt{t}\right )} + \frac14 t^{-1}  \cosh{\left (\sqrt{t}\right )}$$
$$f'''(t) = \frac18 t^{-5/2} (3+t) \sinh{\left (\sqrt{t}\right )} - \frac38 t^{-2} \cosh{\left (\sqrt{t}\right )} $$
Putting this all together, I get that the sum equals

$$\sum_{n=1}^{\infty} \frac{n^2 t^n}{(2 n-1)!} = \frac14 t (1+t) \frac{\sinh{\left (\sqrt{t}\right )}}{\sqrt{t}} + \frac34 t \cosh{\left (\sqrt{t}\right )} $$

A: Suppose $f(t)=\sum_{n} a_n t^n$.  We proceed formally.  Take derivatives, to get $$f'(t)=\sum_n na_n t^{n-1}$$
Multiply by $t$ to get $$tf'(t)=\sum_n na_n t^n$$
Take derivatives again to get $$(tf'(t))'=\sum_n n^2 a_n t^{n-1}$$
Multiply by $t$ again to get the final answer $$t(tf'(t))'=\sum_n n^2 a_n t^n$$
Now, take $f(t)=\sinh t$, and expand out and simplify the LHS of the final answer above.  You may need to adjust a little bit for small $n$.
A: Look at the effect of $\frac d{dt}(t\sinh t)$ on the coefficients of $t^n$ in $\sinh t$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\sum_{n = 1}^{\infty}{n^{2} \over
\pars{2n - 1}!}\,t^{n}} =
{\root{t} \over 4}\sum_{n = 1}^{\infty}{\bracks{\pars{2n - 1} + 1}^{\,2} \over \pars{2n - 1}!}\,\pars{\root{t}}^{2n - 1}
\\[5mm] = &\
{\root{t} \over 4}\sum_{n = 1}^{\infty}{\pars{n + 1}^{\,2} \over n!}
\pars{\root{t}}^{n}\,\,{1 - \pars{-1}^{n} \over 2}
\\[5mm] &\
{\root{t} \over 8}\bracks{%
\sum_{n = 1}^{\infty}{\pars{n + 1}^{\,2} \over n!}
\pars{\root{t}}^{n} -
\sum_{n = 1}^{\infty}{\pars{n + 1}^{\,2} \over n!}
\pars{-\root{t}}^{n}}\label{1}\tag{1}
\end{align}

Note that $\ds{\sum_{n = 1}^{\infty}{x^{n + 1} \over n!} =
x\pars{-1 + \sum_{n = 0}^{\infty}{x^{n} \over n!}} =
-x + x\expo{x}}$ and
\begin{align}
&\left\{\begin{array}{rcl}
\ds{\sum_{n = 1}^{\infty}{n + 1 \over n!}\,x^{n + 1}}
& \ds{=} &
\ds{x\,\totald{\pars{-x + x\expo{x}}}{x}}
\\ & \ds{=} &
\ds{-x + x\expo{x} + x^{2}\expo{x}}
\\[5mm]
\ds{\color{red}{\sum_{n = 1}^{\infty}{\pars{n + 1}^{2} \over n!}\,x^{n}}} & \ds{=} &
\ds{\totald{\pars{-x + x\expo{x} + x^{2}\expo{x}}}{x}}
\\ & \ds{=} &
\ds{\color{red}{\pars{x^{2} + 3x + 1}\expo{x} - 1}}
\end{array}\right.
\end{align}

Then ( see (\ref{1}) ),
\begin{align}
&\bbox[5px,#ffd]{\sum_{n = 1}^{\infty}{n^{2} \over
\pars{2n - 1}!}\,t^{n}}
\\[5mm] = &\
{\root{t} \over 8}\bracks{%
\pars{t + 3\root{t} + 1}\expo{\root{t}} - 1}
\\[2mm] - &\
{\root{t} \over 8}\bracks{%
\pars{t - 3\root{t} + 1}\expo{-\root{t}} - 1}
\\[5mm] = &\
\bbx{{1 \over 4}\root{t}\pars{t + 1}\sinh\pars{\root{t}} +
{3 \over 4}t\cosh\pars{\root{t}}} \\ &
\end{align}
