<s>[INST] Will the series $\sum\limits_{n=1}^\infty \frac{n^2-n-1}{n!}$ converge when it runs from 1 to infinity? $$\sum_{n=1}^\infty \frac{n^2-n-1}{n!}$$
I usually divide by $n$ or $n^2$ but I'm unable to do so because of the factorial term. 
 [/INST] The series evaluation:
\begin{align}
\sum_{n=1}^{\infty} \frac{n^{2} - n - 1}{n!} &= \sum_{n=1}^{\infty} \left(\frac{1}{(n-2)!} - \frac{1}{n!}\right) \\
&= \sum_{n=2}^{\infty} \frac{1}{(n-2)!} - (e^{1} -1) \\
&= \sum_{n=0}^{\infty} \frac{1}{n!} - (e^{1} - 1) = e^{1} - (e^{1} - 1) = 1.
\end{align}
Convergence by ratio test:
$$\lim_{n \to \infty} \frac{(n+1)^2 - (n+1) - 1}{(n+1)!} \cdot \frac{n!}{n^2 - n -1} = \lim_{n \to \infty} \frac{1}{n+1} \cdot \frac{1 + \frac{1}{n} - \frac{1}{n^{2}}}{1 - \frac{1}{n} - \frac{1}{n^{2}}} = 0.$$
Since the limit is less than one the series converges.
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