Proof that $1-\frac{1}{e}\sum_{k=0}^{n} \frac{1}{k!} \leq \frac{1}{n!}$ for $n \in \mathbb{N}^*$ For $n \in \mathbb{N}^*$, consider the function $f_n(x)=e^{-x}\sum_{k=0}^{n} \frac{x^k}{k!}$ over $[0;1]$.
I have to prove the following inequality : $\forall n \in \mathbb{N}^*:0 \leq f_n(0)-f_n(1) \leq \frac{1}{n!}$. By taking the derivative and showing the function is decreasing over $[0;1]$, I am able to prove the LHS, however I struggle to prove the RHS... How would you do?
Edit: The point of this question is to then prove the convergence of the infinite serie $\sum_{k=0}^{\infty} \frac{1}{k!}$ to $e$, which means such a result is not supposed to be used.
 A: Proof without induction :
One has $$e- \sum_{k=0}^n \dfrac{1}{k!} = \sum_{k=n+1}^{+\infty} \dfrac{1}{k!} = \sum_{k=0}^{+\infty} \dfrac{1}{(k+n+1)!} \quad \quad (*)$$
But for every $k \geq 0$, one has $$\dfrac{(k+n+1)!}{(n+1)!k!} = {k+n+1 \choose n+1} \geq 1, \quad \text{so} \quad \dfrac{1}{(k+n+1)!} \leq \dfrac{1}{(n+1)!k!}$$
Then $(*)$ becomes $$e- \sum_{k=0}^n \dfrac{1}{k!} \leq \sum_{k=0}^{+\infty} \dfrac{1}{(n+1)!k!} = \dfrac{e}{(n+1)!}, \quad \text{i.e.} \quad \boxed{1- \dfrac{1}{e}\sum_{k=0}^n \dfrac{1}{k!} \leq \dfrac{1}{(n+1)!}}$$
(which is a little bit more precise than the inequality you want to prove).
A: The most straightforward proof uses Lagrange error bound for the Maclaurin series for $x \mapsto e^x$.
This bound shows us that $|e - \sum\limits_{k = 0}^n \frac{1}{k!}| \leq \frac{e}{(n+ 1)!}$. We can drop the absolute value sign to conclude $e - \sum\limits_{k = 0}^n \frac{1}{k!} \leq \frac{e}{(n+ 1)!}$.
Then dividing both sides by $e$, we have $1 - \frac{1}{e} \sum\limits_{k = 0}^n \frac{1}{k!} \leq \frac{1}{(n + 1)!} \leq \frac{1}{n!}$.
A: I think I got the most elementary proof of the result. We want to prove that
$e-\sum_{k=0}^{n}\dfrac{1}{k!}\leq e\dfrac{1}{n!}$
which is,  $\,\,\,\sum_{k=n+1}^{+\infty}\dfrac{n!}{k!}\leq\,e\,\,$ i.e.
$\dfrac{1}{n+1}+\dfrac{1}{(n+1)(n+2)}+\dfrac{1}{(n+1)(n+2)(n+3)}+...\leq\,e$.
We observe that $\dfrac{1}{(n+1)(n+2)}\leq\,\dfrac{1}{(n+1)^{2}} , \dfrac{1}{(n+1)(n+2)(n+3)}\,\leq\,\dfrac{1}{(n+1)^{3}},$ e.t.c.
so we obtain a geometric series $\dfrac{1}{n+1}(\dfrac{1}{1-\dfrac{1}{n+1}})$=$\dfrac{1}{n}\,\leq\,e$  and the result is proved!!
A: Taking the derivative, we have
$$
f_n'(x) = - e^{-x} \sum_{k = 0}^n \frac{x^k}{k!} + e^{-x} \sum_{k=1}^n \frac{x^{k-1}}{(k-1)!} = - e^{-x} \sum_{k = 0}^n \frac{x^k}{k!} + e^{-x} \sum_{k = 1}^{n-1} \frac{x^k}{k!} = -e^{-x} \frac{x^n}{n!} \leq 0
$$
for any $x \in [0,1]$.
Thus, as you noticed, the function is decreasing in $[0,1]$, and you get $f_n(0) \geq f_n(1)$.
Now, from the mean-value theorem there exists $c \in [0,1]$ such that
$$
f_n(0) - f_n(1) = -f_n(c)' = e^{-c} \frac{c^n}{n!} \leq e^{-1} \frac{1}{n!} \leq \frac{1}{n!}
$$
Edit: For completeness, the inequality $e^{-c} \frac{c^n}{n!} \leq e^{-1} \frac{1}{n!}$ follows from the fact that $g(x) = e^{-x} \frac{x^n}{n!}$ is an increasing function for $x \in [0,1]$, as
$$
g'(x) = -e^{-x} \frac{x^n}{n!} +  e^{-x} \frac{x^{n-1}}{(n-1)!} = e^{-x}x^{n-1}\frac{n - x}{n!} \geq 0
$$
