Find values for the given sums

Given the fact that

$\sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x \qquad \text{for all } x.$

Assuming that we can operate on infinite sums in the same way we can operate on finite sums, find expressions for each of the following sums.

We compute,
$\begin{align*} \sum_{n=2}^{\infty} \frac{n-1}{n!} &= \sum_{n=2}^{\infty} \frac{1}{(n-1)!} - \sum_{n=2}^{\infty} \frac{1}{n!} \\[9pt] &= \sum_{n=1}^{\infty} \frac{1}{n!} - \sum_{n=2}^{\infty} \frac{1}{n!} \\[9pt] &= \sum_{n=1}^{\infty} \frac{1}{n!} - \sum_{n=1}^{\infty} \frac{1}{n!} + 1 \\[9pt] &= 1. \end{align*}$
We compute,
$\begin{align*} \sum_{n=2}^{\infty} \frac{n+1}{n!} &= \sum_{n=2}^{\infty} \frac{1}{(n-1)!} + \sum_{n=2}^{\infty} \frac{1}{n!} \\[9pt] &= \sum_{n=1}^{\infty} \frac{1}{n!} + \sum_{n=2}^{\infty} \frac{1}{n!} \\[9pt] &= 2 \sum_{n=1}^{\infty} \frac{1}{n!} - 1 - 2 \\[9pt] &= 2e - 3. \end{align*}$
We compute,
$\begin{align*} \sum_{n=2}^{\infty} \frac{(n-1)(n+1)}{n!} &= \sum_{n=2}^{\infty} \frac{(n+1)}{n (n-2)!} \\[9pt] &= \sum_{n=2}^{\infty} \frac{1}{(n-2)!} + \sum_{n=2}^{\infty} \frac{1}{n(n-2)!} \\[9pt] &= e + \sum_{n=2}^{\infty} \frac{n-1}{n!} \\[9pt] &= e + 1 &(\text{part (a)}). \end{align*}$

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