prove :$ \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}< \frac{5}{2}-\frac{1}{n},\forall n\in \mathbb{N}^{*}$ prove :$ \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}< \frac{5}{2}-\frac{1}{n},\forall n\in \mathbb{N}^{*}$
i tried to do this with induction so i d like to know if what i did is correct:
for $n=1$  the statement is true since $1<\frac{5}{2}-1$ 
if the statement is true for for n we have :
 $\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}+\frac{1}{n+1!}< \frac{5}{2}-\frac{1}{n+1},\forall n\in \mathbb{N}^{*}$
which is true since as n gets larger the sum increases at a lower rate whereas the difference increases at a higher rate because the fraction gets smaller and smaller... so because the statement stands for n it will also stand for $n + 1$.
 A: One can use the series of $e$. For $n=1$, your statement is true. Let us deal with $n\ge 2$.
We have that
$$\sum_{n\ge 1} \frac{1}{n!} = e-1\,.$$
Then the claim follows if we prove that
$$ e\le \frac{7}{2}-\frac{1}{n}\,, \qquad \forall n\ge 2\,.$$
As the LHS is less than $3$ and the RHS is greater or equal than $3$ the claim follows.

Your approach follows another idea, which is good. You need to put it down more formally. By induction hypothesis you have
$$\frac{1}{1!}+\dots \frac{1}{n!}+\frac{1}{(n+1)!} < \frac 52 - \frac 1n +\frac{1}{(n+1)!}\,.$$
If you can prove that
$$- \frac 1n +\frac{1}{(n+1)!} < -\frac{1}{n+1}\,,$$
then you are done. Now it's up to you!
A: The correct way to do the inductive step is
$$\begin{align}
{1\over1!}+{1\over2!}+\cdots+{1\over n!}+{1\over(n+1)!}
&\le{5\over2}-{1\over n}+{1\over(n+1)!}\\
&\le{5\over2}-{1\over n}+{1\over(n+1)n}\\
&={5\over2}-{1\over n+1}
\end{align}$$
The first inequality is the inductive hypothesis, the second inequality follows from $(n+1)!=(n+1)n(n-1)\cdots2\cdot1\ge(n+1)n$ for $n\ge1$, and the final equality follows from
$${1\over(n+1)n}-{1\over n}={1\over(n+1)n}-{(n+1)\over(n+1)n}={-n\over(n+1)n}={-1\over n+1}$$
A: Hint: Show that for $n\ge1$, we have $\frac1{(n+1)!}\le\frac1n-\frac1{n+1}=\frac1{n(n+1)}$.
Then attempt the induction.

Another approach is to show by a very easy induction that for $k\ge1$, $k!\ge2^{k-1}$. Then show that
$$
\sum_{k=1}^n\frac1{k!}\le\sum_{k=1}^n\frac1{2^{k-1}}\lt2
$$
Furthermore, note that for $n\ge2$, we have $\frac52-\frac1n\ge2$.
You'll have to do the case $n=1$ by hand, however.
