logarithms - Prove that $5/2 < e < 3$?

Prove that $$\dfrac{5}{2} < e < 3$$

By the definition of $\log$ and $\exp$,
$$1 = \log(e) = \int_1^e \dfrac{1}{t} dt$$

where $e = \exp(1)$.

So given that $e$ is unknown, how could I prove this problem? I think I'm missing some important facts that could somehow help me rewrite $e$ in some form of $3$ and $5/2$. Any idea would be greatly appreciated.

Answer

$e=\lim_{n\to \infty}(1+\frac1n)^n$

the rth term $t_r=\frac{n(n-1)(n-2)\cdots(n-r+1)}{r!n^r}=\frac1{r!}\prod_{0\le s

So, $\lim_{n\to \infty}t_r=\frac1{r!}$

So, $e=1+\frac1{1!}+\frac1{2!}+\frac1{3!}+\cdots$

But $1+\frac1{1!}+\frac1{2!}+\frac1{3!}+\cdots>1+1+0.5=2.5$

Again,

$3!=1.2.3>1.2.2=2^2$

$4!=1.2.3.4>1.2.2.2=2^3$

So,

$e=1+\frac1{1!}+\frac1{2!}+\frac1{3!}+\cdots$
$<1+1+\frac12+\frac1{2^2}+\frac1{2^3}+\cdots$
$=1+(1+\frac12+\frac1{2^2}+\frac1{2^3}+\cdots)=1+\frac{1}{1-\frac12}=3$
as the terms inside parenthesis forms an infinite geometric series with the common ratio $=\frac12,$ the 1st term being $1$