How the inequality $\frac1{n+2} \leq \left|\frac{n!}{e}-!n\right|$ holds? $\begin{align*}\frac{n!}{e}-!n&=n!\sum_{k=n+1}^\infty \frac{(-1)^k}{k!}\\&=\frac{(-1)^{n+1}}{n+1}\left(1-\frac1{n+2}+\frac1{(n+2)(n+3)}-\cdots\right)\end{align*}$
and hence
$$\frac1{n+2} \leq \left|\frac{n!}{e}-!n\right| \leq \frac1{n+1}$$
I can understand how$$ \left|\frac{n!}{e}-!n\right| \leq \frac1{n+1}$$
is but can't understand how the inequality $$\frac1{n+2} \leq \left|\frac{n!}{e}-!n\right| $$ is true.
Please HELP.
Thank you.
 A: I think $$\frac1{n+2} \leq \left|\frac{n!}{e}-!n\right|$$ is because $$\frac{1}{n+1}-\frac{1}{(n+1)(n+2)}=\frac{1}{n+2}$$ and there are many many terms to add in $\frac{1}{n+2}$ so it is less than $\left|\frac{n!}{e}-!n\right|$.
Is my answer correct?
Please verify!
Thank you.
A: Your derivation and your answer are correct, and only one minor piece of detail is helpful in guaranteeing the result:
$${1\over n+2}\le |{n!\over e}-!n|\le {1\over n+1}\tag{1}$$
The far LHS of $(1)$ is shown by the following condition:
$${1\over n+1}-{1\over (n+1)(n+2)}+{1\over (n+1)(n+2)(n+3)}-\dots$$
$$=\left({1\over n+1}-{1\over (n+1)(n+2)}\right)+\left({1\over (n+1)(n+2)(n+3)}-{1\over (n+1)(n+2)(n+3)(n+4)}\right)\dots$$
$$={1\over (n+2)}+{1\over (n+1)(n+2)(n+4)}+\dots\ge {1\over n+2}$$
The argument is very similar for the far RHS of $(1)$, except that the sign of the terms in the sum is switched due to the sum being grouped after the first term instead of immediately.
The grouping of terms is a viable technique as all sums are absolutely convergent.
The absolute value, of course, takes care of the $(-1)^{n+1}$ in the way that you pulled it out very effectively.
