$S=\sum_{k=2}^{n}\frac{k^{2}-2}{k!}, n\geq 2$

Question

$$S=\sum_{k=2}^{n}\frac{k^{2}-2}{k!}, n\geq 2$$

I got $S=\sum_{k=2}^{n}\frac{1}{(k-2)!}+\frac{1}{(k-1)!}-\frac{1}{k!}-\frac{1}{k!}$

I give k values but not all terms are vanishing.I remain with $\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{(n-2)!}$

The sum should be $2-e+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{(n-2)!}$

Answer 1

For the sum until the index $n$, you can apply the same trick : \begin{align*} \sum_{k=2}^n \frac{k^2-2}{k!} &= \sum_{k=2}^n \left[\frac{k (k-1)}{k!} + \frac{k}{k!}-\frac{2}{k!} \right]\\ &= \color{blue}{\sum_{k=2}^n \frac{k (k-1)}{k!}} + \color{orange}{\sum_{k=2}^n\frac{k}{k!}}-\color{green}{\sum_{k=2}^n\frac{2}{k!}}\\ &= \color{blue}{\sum_{k=0}^{n-2} \frac{1}{k!}} + \color{orange}{\sum_{k=0}^{n-2}\frac{1}{k!} - 1 + \frac{1}{(n-1)!}}- \color{green}{2\sum_{k=0}^{n-2}\frac{1}{k!} + 2+2-\frac{2}{n!}-\frac{2}{(n-1)!}}\\ &= 3-\frac{2}{n!}-\frac{1}{(n-1)!} \end{align*}

When $n$ tends to $\infty$, this converges to $3$.

Answer 2

$$ \eqalign{ & \sum\limits_{k = 2}^n {\left( {{1 \over {\left( {k - 2} \right)!}} + {1 \over {\left( {k - 1} \right)!}} - {2 \over {k!}}} \right)} = \sum\limits_{k = 0}^{n - 2} {{1 \over {k!}} + \sum\limits_{k = 1}^{n - 1} {{1 \over {k!}}} - \sum\limits_{k = 2}^n {{2 \over {k!}}} } = \cr & = 1 + 1 + \sum\limits_{k = 2}^{n - 2} {{1 \over {k!}}} + 1 + \sum\limits_{k = 2}^{n - 2} {{1 \over {k!}}} + {1 \over {\left( {n - 1} \right)!}} - 2\left( {\sum\limits_{k = 2}^{n - 2} {{1 \over {k!}}} + {1 \over {\left( {n - 1} \right)!}} + {1 \over {n!}}} \right) = \cr & = 3 - {1 \over {\left( {n - 1} \right)!}} - {2 \over {n!}} = 3 - {{n + 2} \over {n!}} \cr} $$ and it checks with the original sum, which in fact is rational and cannot include $e$.
So something is wrong somewhere in your notes.