<s>[INST] 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.
 [/INST] Method 1:
Using the power series for $\log(1+x)$, we get
$$
\begin{align}
\log\left(1+\frac1n\right)
&=-\log\left(1-\frac1{n+1}\right)\\
&=\frac1{n+1}+\frac1{2(n+1)^2}+\frac1{3(n+1)^3}+\dots\tag{1}
\end{align}
$$
Multiply $(1)$ by $n+1$:
$$
(n+1)\log\left(1+\frac1n\right)=1+\frac1{2(n+1)}+\frac1{3(n+1)^2}+\dots\tag{2}
$$
$(2)$ is obviously decreasing in $n$, therefore $\displaystyle\left(1+\frac1n\right)^{\large n+1}$ is decreasing in $n$.
Multiply $(1)$ by $n$:
$$
\begin{align}
n\log\left(1+\frac1n\right)
&=((n+1)-1)\log\left(1+\frac1n\right)\\
&=1-\frac1{1\cdot2(n+1)}-\frac1{2\cdot3(n+1)^2}-\dots\tag{3}
\end{align}
$$
$(3)$ is obviously increasing in $n$, therefore $\displaystyle\left(1+\frac1n\right)^{\large n}$ is increasing in $n$.
Therefore, since $\displaystyle e=\lim_{n\to\infty}\left(1+\frac1n\right)^{\large n}$, we have
$$
\left(1+\frac1n\right)^{\large n}< e<\left(1+\frac1n\right)^{\large n+1}\tag{4}
$$
Let $n=6$, then
$$
\frac52<\left(\frac76\right)^{\large 6}< e<\left(\frac76\right)^{\large 7}<3\tag{5}
$$
Method 2:
Since $\displaystyle e=\sum_{n=0}^\infty\frac1{n!}\ $, we have
$$
\begin{align}
\frac52=1+1+\frac12< e
&=1+\frac11+\frac1{1\cdot2}+\frac1{1\cdot2\cdot3}+\frac1{1\cdot2\cdot3\cdot4}+\dots\\
&<1+\frac11+\frac1{1\cdot2}+\frac1{1\cdot2\cdot2}+\frac1{1\cdot2\cdot2\cdot2}+\dots\\
&=3\tag{6}
\end{align}
$$
</s>