Vtgyjfy

Let $S_n$ be a sequence defined by:
$$
S_n = 1 + sum_{k=1}^n {1over k!}
$$
Prove that:
$$
e - S_n le frac{n+2}{n!(n+1)^2}
$$

This problem comes in the limits section so i may use anything before the definition of a derivative. I've started with the following. Define:
$$
L_n = left(1+{1over n}right)^n
$$

Rewrite $L_n$ as follows:
$$
begin{align}
L_n &= left(1+{1over n}right)^n \
&= 1+ sum_{k=1}^n {nchoose k}{1over n^k} \
&= 1 + sum_{k=1}^nfrac{overbrace{ncdot(n-1)cdot(n-2)cdots(n-(k-1))}^{n - k + 1
text{ times}}}{k!n^k}
end{align}
$$

I've then tried to consider the difference between $L_n$ and $S_n$:
$$
L_n - S_n = {1over 2!}left(1-{1over n}right) - {1over 2!} + {1over 3!}left(1-{1over n}right)left(1-{2over n}right) - {1over 3!} + cdots \
= {1over 2!}left(1-{1over n} - 1right) + {1over 3!}left(left(1-{1over n}right)left(1-{2over n}right) - 1right) + cdots
$$

Consider each parentheses. I've tried defining a sequence such that:
$$
forall n ge 2:a_n = {1over n!}left(prod_{k=1}^{n-1}left(1-{kover n}right) - 1right)
$$

This sequence seems to always be less than $0$. So we have that:
$$
forall nin Bbb N :a_n le 0
$$

Since $a_n$ is involved in $L_n - S_n$ we may also conclude that it is also less than $0$, namely:
$$
L_n - S_n = sum_{k=2}^na_n le 0
$$

But at the same time:
$$
frac{n+2}{n!(n+1)^2} ge 0
$$

So by this we have that:
$$
L_n - S_n le e - S_n le 0 le frac{n+2}{n!(n+1)^2}
$$

Now I don't see how to proceed. Should I bound the difference from another side? It feels like all i've done so far (above) is not even a usable argument for this problem, so could you please help me prove what's in the problem statement or point to the right direction.

I've made a Visualization of what is written above, perhaps that would be helpful.

We will use the following limit. If $S_n=sumlimits_{k=0}^{n}frac{1}{k!}$ then
$$
limlimits_{nrightarrowinfty} S_n = e.
$$
This limit can be derived from $limlimits_{nrightarrowinfty} left(1+frac{1}{n}right)^n = e$.

Now we will prove that for any $mgeq 1$ inequality $S_{n+m}leq S_n+frac{n+2}{(n+1)(n+1)!}$ holds. Indeed,
$$
S_{n+m}-S_n=frac{1}{(n+1)!}+frac{1}{(n+2)!}+ldots+frac{1}{(n+m)!}leq frac{1}{(n+1)!}left(1+frac{1}{n+2}+ldots+frac{1}{(n+2)^{m-1}}right)=frac{1}{(n+1)!}cdot frac{1-frac{1}{(n+2)^m}}{1-frac{1}{n+2}}<frac{n+2}{(n+1)(n+1)!}.
$$

From inequality $S_{n+m}leq S_n+frac{n+2}{(n+1)(n+1)!}$ when $m$ tend to infinity we get $eleq S_n+frac{n+2}{(n+1)(n+1)!}$ as desired.

$begin{array}\
e-S_n
&= sum_{k=n+1}^{infty} {1over k!}\
&= frac1{(n+1)!}sum_{k=n+1}^{infty} {(n+1)!over k!}\
&= frac1{(n+1)!}sum_{k=0}^{infty} {(n+1)!over (n+1+k)!}\
&= frac1{(n+1)!}left(1+sum_{k=1}^{infty} {(n+1)!over (n+1+k)!}right)\
&= frac1{(n+1)!}left(1+sum_{k=1}^{infty} {1over prod_{j=1}^{k}(n+1+j)}right)\
&< frac1{(n+1)!}left(1+sum_{k=1}^{infty} {1over prod_{j=1}^{k}(n+2)}right)\
&= frac1{(n+1)!}left(1+sum_{k=1}^{infty} {1over (n+2)^k}right)\
&= frac1{(n+1)!}left(1+frac{1/(n+2)}{1-1/(n+2)}right)\
&= frac1{(n+1)!}left(1+frac{1}{n+1}right)\
&= frac1{(n+1)!}left(frac{n+2}{n+1}right)\
&= frac{n+2}{(n+1)(n+1)!}\
&= frac{n+2}{(n+1)^2n!}\
end{array}
$