<s>[INST] Sum $\sum_{n=1}^\infty \frac{n^2}{(n+2)!}$ Problem is to find sum
$$\frac{1}{3!}+\frac{4}{4!}+\frac{9}{5!}\cdots$$
What I knew doesn't apply on this problem
Some series are telescoping, some types are solvable using binomial , both look useless here
Binomial gives
$$n (n-1) (n-2)x^3=1$$
 $$n (n-1)( n-2)(n-3)x^4=4$$
What approach to use here?
 [/INST] Replace the range so that it starts at $n = 0$ and substitute $n^2 = (n+2)(n+1) - 3(n+2) + 4$ to obtain
$$
\begin{align*}
\sum_{n=0}^\infty \frac{n^2}{(n+2)!} &=
\sum_{n=0}^\infty \frac{(n+2)(n+1)}{(n+2)!} -
\sum_{n=0}^\infty \frac{3(n+2)}{(n+2)!} +
\sum_{n=0}^\infty \frac{4}{(n+2)!} \\ &=
\sum_{n=0}^\infty \frac{1}{n!} -
3\sum_{n=0}^\infty \frac{1}{(n+1)!} +
4\sum_{n=0}^\infty \frac{1}{(n+2)!} \\ &=
\sum_{n=0}^\infty \frac{1}{n!} -
3\sum_{n=1}^\infty \frac{1}{n!} +
4\sum_{n=2}^\infty \frac{1}{n!} \\ &=
(1-3+4) \sum_{n=0}^\infty \frac{1}{n!} -3\left(-\frac{1}{0!}\right) +4\left(-\frac{1}{0!}-\frac{1}{1!}\right) \\ &=
2e+3-8 \\ &= 2e-5.
\end{align*}
$$
More generally, this shows that for every polynomial $P(n)$ over the rationals and $k \geq 0$, we have
$$
\sum_{n=0}^\infty \frac{P(n)}{(n+k)!} = \alpha e + \beta
$$
for some rationals $\alpha,\beta$, and these can be calculated by presenting $P(n)$ as a linear combination of the polynomials
$$ 1, n+k, (n+k)(n+k-1), (n+k)(n+k-1)(n+k-2), \ldots. $$
In your case, we needed to represent $n^2$ as a linear combination of $1,n+2,(n+2)(n+1)$. The general form of such a combination is
$$ n^2 = A+B(n+2)+C(n+2)(n+1) = Cn^2 + (3C+B)n + (2C+2B+A). $$
Equating coefficients, we get the linear system $C = 1$, $3C+B = 0$, $2C+2B+A = 0$, which we can solve by back-substitution to obtain $C=1$, $B=-3$, $A=4$. Back-substitution works in general, and we obtain a simple and efficient algorithm for evaluating the general form of this series.
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