Some infinite series

[gugo82]

Problem:

Find the sum of the series:
\[
\begin{align}
& \sum_{n=0}^\infty \frac{n^4}{n!} \\
& \sum_{n=0}^\infty \frac{(n+1)^2}{n!}\\
& \sum_{n=0}^\infty \frac{(n+1)^3}{n!}\; .
\end{align}
\]

[Studente Anonimo]

Since \( (1) = (3)\) (after a simplification and a change of variable), we focus on \( (2) \) and \( (3)\).

\( (2) . \) We have that \[ \sum_{n = 0}^\infty \frac{(n+1)^2}{n!} = \sum_{n = 0}^\infty \frac{n^2}{n!} + \frac{2n}{n!} + \frac{1}{n!}. \]Since everything here is absolutely convergent we can harmlessly permute the summation order: \[ \sum_{n = 0}^\infty \frac{n}{n!} = \sum_{n = 1}^\infty \frac{1}{(n-1)!} = e \] and \[\sum_{n = 0}^\infty \frac{n^2}{n!} = \sum_{n = 1}^\infty \frac{n}{(n-1)!} = \sum_{n=0}^\infty \frac{n+1}{n!} = 2e. \] In conclusion the sum of \( (2) \) equals \(5 e \).

\( (3). \) With similar computations it turns out that the sum of \( (3) \) equals \( 15 e\).

I would like to suggest a problem, too:

Using \( \sum_{n=1}^\infty \frac{1}{n^2}= \frac{\pi^2}{6} \), compute \[ \sum_{n=1}^\infty \frac{1}{(n+2)^2} \left( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n \cdot (n+1)}\right). \]

[gugo82]

"gugo82":Problem:

Find the sum of the series:
\[ \begin{align} & \sum_{n=0}^\infty \frac{n^4}{n!} \\ & \sum_{n=0}^\infty \frac{(n+1)^2}{n!}\\ & \sum_{n=0}^\infty \frac{(n+1)^3}{n!}\; . \end{align} \]

(1) We can rewrite:
\[
\sum_{n=0}^\infty \frac{n^4}{n!} = \sum_{n=1}^\infty \frac{n^3}{(n-1)!} = \sum_{n=0}^\infty \frac{(n+1)^3}{n!}
\]
hence the sum of (1) equals the sum of (3), which will be evaluated in a while.

(2) We have:
\[
\sum_{n=0}^\infty \frac{(n+1)^2}{n!} = \sum_{n=0}^\infty \frac{n^2 + 2n +1}{n!} = \sum_{n=0}^\infty \frac{n(n-1) +3n + 1}{n!} = \sum_{n=2}^\infty \frac{1}{(n-2)!} + 3 \sum_{n=1}^\infty \frac{1}{(n-1)!} + \sum_{n=0}^\infty \frac{1}{n!} = e + 3e + e = 5e\;.
\]

(3) We find:
\[
\sum_{n=0}^\infty \frac{(n+1)^3}{n!} = \sum_{n=0}^\infty \frac{n^3 + 3n^2 + 3n +1}{n!} = \sum_{n=0}^\infty \frac{n(n-1)(n-2) + 6n(n-1) + 7n +1}{n!} = \sum_{n=3}^\infty \frac{1}{(n-3)!} + 6 \sum_{n=2}^\infty \frac{1}{(n-2)!} + 7 \sum_{n=1}^\infty \frac{1}{(n-1)!} + \sum_{n=0}^\infty \frac{1}{n!} = e + 6e + 7e + e = 15e\; .
\]

***

Addendum:

Is it possibile to write down an explicit formula for $sum_(n=0)^oo (n+k)^2/(n!)$ and $sum_(n=0)^oo (n+k)^3/(n!)$ (with $k in NN$)?
And what about a formula for $sum_(n=0)^oo (n+1)^h/(n!)$ (with $h in NN$)?