Recently, when I self-learnt Discrete Mathematics and Its Applications 8th by Kenneth Rosen, I found something with confusion.
In the description of the probability of the derangement (proofwiki link)
Because this is an alternating series with terms tending to zero, it follows that as n grows without bound, the probability that no one receives the correct hat converges to $e^{−1} \approx 0.368$. In fact, this probability can be shown to be within $1/(n + 1)!$ of $e^{−1}$.
Then based on the Note in the proofwiki link, when taking the error in account: $$ \begin{align*} \sum_{k=0}^{n}\frac{(-1)^k}{k!}&=\sum_{k=0}^{\infty}\frac{(-1)^k}{k!}-\sum_{k=n+1}^{\infty}\frac{(-1)^k}{k!}\\ &=\sum_{k=0}^{\infty}\frac{(-1)^k}{k!}-\sum_{m=0}^{\infty}(\frac{(-1)^{n+1}}{(m+n+1)^{\underline{n+1}}}\cdot\frac{(-1)^m}{m!})\tag{$\ast$} \end{align*} $$
Q:
Then is there one way to get $(\ast)<\frac{e^{-1}}{(n + 1)!}$? Or if my calculation is wrong, how to get the conclusion "within $1/(n + 1)!$ of $e^{−1}$"?
Using https://en.wikipedia.org/wiki/Alternating_series#Approximating_sums, the error in approximating $e^{-1}=\sum_{k\ge 0}(-1)^k/k!$ by the partial sum $\sum_{k=0}^n (-1)^k/k!$ is at most equal to the absolute value of the next term in the sequence, which is $|(-1)^{n+1}/(n+1)!|=1/(n+1)!$. That is, $$\left|e^{-1}-\sum_{k=0}^n (-1)^k/k!\right|\le \frac1{(n+1)!},$$which is exactly what you want.