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.

Let $$e_n=\left(1+\frac{1}{n}\right)^n,\quad s_n=\sum_{k=0}^{n}\frac{1}{k!}\ (0!=1),$$
then $e_1=2=s_1,e_2=\frac{9}{4}<\frac{5}{2}=s_2$, when $n>2$,
\begin{align*}
\left(1+\frac{1}{n}\right)^n
 &=1+1+\sum_{k=2}^{n}\frac{1}{k!}\left(1-\frac{1}{n}\right)\cdots\left(1-\frac{k-1}{n}\right)\\
 &<1+1+\sum_{k=2}^{n}\frac{1}{k!}\\
 &=1+1+\frac{1}{2}+\sum_{k=3}^{n}\frac{1}{k!}\\
 &\leq\frac{5}{2}+\sum_{k=3}^{n}\frac{1}{k(k-1)(k-2)}\\
 &=\frac{5}{2}+\frac{1}{2}\sum_{k=3}^{n}\frac{1}{k-1}\left(\frac1{k-2}-\frac1k\right)\\
 &=\frac{5}{2}+\frac{1}{2}\sum_{k=3}^{n}\frac{1}{(k-1)(k-2)}-\frac{1}{2}\sum_{k=3}^{n}\frac{1}{k(k-1)}\\
 &=2.75+\frac{1}{2}\left(\frac1n-\frac{1}{n-1}\right)\\
 &<2.75.
\end{align*}
Let $n\to \infty$, we have $$e\leq 2.75<3.$$
