How to calculate the sum of a sequence of reciprocal of factorial I wonder how to calculate the sum of the below sequence
\begin{align}
S & = \frac{1}{2!} - \frac{2}{3!} + \frac{3}{4!} - \frac{4}{5!} + \frac{5}{6!} - ... \\
& = \frac{2 - 1}{2!} - \frac{3 - 1}{3!} + \frac{4 - 1}{4!} - \frac{5 - 1}{5!} + ...\\
& = 1 - \frac{1}{2!} - \frac{1}{2!} + \frac{1}{3!} + \frac{1}{3!} - \frac{1}{4!} - \frac{1}{4!} + \frac{1}{5!}...\\
& = 1 - 2 * (\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}...）\\
& = ？
\end{align}
There is a similar sequence, which is easy to solve
\begin{align}
S & = \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \frac{4}{5!} + \frac{5}{6!} + ... \\
& = \frac{2 - 1}{2!} + \frac{3 - 1}{3!} + \frac{4 - 1}{4!} + \frac{5 - 1}{5!} + ...\\
& = 1 - \frac{1}{2!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{3!} - \frac{1}{4!} + \frac{1}{4!} - \frac{1}{5!}...\\
& = 1 - \frac{1}{n!}
\end{align}

$$\sum_{n!=2}^\infty (-1)^{n-1}\frac1{n!} = 1 - \frac1e.$$
