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

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

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

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 $1$

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+\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+\frac{1}{2}+\frac{1}{6}+\frac{1}{8}\Rightarrow e\le 1+\frac{1}{2}+\frac{1}{6}+\frac{1}{8}<3 \end{equation}

From $$\log(x) = \int_1^x\frac{dt}{t}$$ and using a constant lower and upper bound for $1/t$ on the interval $[1, x]$ it follows that $$1 - \frac{1}{x} \leq \log(x) \leq x - 1$$ for all $x > 0$. Taking inverse functions this becomes $$1 +x \leq e^x \leq \frac{1}{1-x}$$ for all $x < 1$ (the lower bound holds for all $x \in \mathbb{R}$, you might want to draw a picture). Take $n \geq 1$, substitute $x \leftarrow x/n$ and raise to the power $n$ to get $$\left(1 + \frac{x}{n}\right)^n \leq e^{\frac{x}{n}n} = e^x \leq \left(1 - \frac{x}{n}\right)^{-n}$$ for all $x < n$. For $x=1$ and $n=6$ this becomes

$$\frac{5}{2} < \left(1 + \frac{1}{6}\right)^6 \leq e \leq \left(1-\frac{1}{6}\right)^{-6} < 3.$$