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.

As we know $e=\lim_{n\to +\infty}\left(1+\dfrac{1}{n}\right)^n$. That sequence is increasing (why?) and so $$\frac{5}{2}<\left(1+\dfrac{1}{10}\right)^{10}<e$$
In addition, for  $n=1,2...$, by the Binomial Theorem,
\begin{gather}\left(1+\dfrac{1}{n}\right)^n=\sum_{k=0}^{n}\binom{n}{k}\left(\dfrac{1}{n}\right)^k=\sum_{k=0}^{n}\frac{1}{k!}\frac{n!}{(n-k)!}\frac{1}{n^k}=
1+\sum_{k=1}^{n}\frac{1}{k!}\frac{n(n-1)...(n-k+1)}{n^k}= \notag\\
1+\sum_{k=1}^{n}\frac{1}{k!}\frac{n}{n}\frac{n-1}{n}...\frac{n-k+1}{n}=
1+\sum_{k=1}^{n}\frac{1}{k!}\left(1-\frac{1}{n}\right)...\left(1-\frac{k-1}{n}\right)<1+\sum_{k=1}^{n}\frac{1}{k!}\end{gather}
(you can skip the above if you know $e=\sum_{n=0}^{\infty}\frac{1}{n!}$)
Since $k\ge 4\Rightarrow k!\ge 2^{k}$,
\begin{equation}\left(1+\dfrac{1}{n}\right)^n<1+1+\frac{1}{2}+\frac{1}{6}+\sum_{k=4}^{n}\frac{1}{2^{k}}\end{equation}
For the last sum, observe that
\begin{equation}\sum_{k=4}^{n}\frac{1}{2^{k}}=\frac{1}{16}\frac{\frac{1}{2^{n-4}}-1}{\frac12-1}=\frac{1}{8}-\frac{1}{2^{n-1}}<\frac{1}{8}
\end{equation}
and thus,
\begin{equation}\left(1+\dfrac{1}{n}\right)^n<1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{8}\Rightarrow e\le 1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{8}<3
\end{equation}
