Proving this inequality $\left(1+\frac{1}{n}\right)^n < 3$

Have a look at page 99 of these notes, where I proved the sharper inequality $\left(1+\frac{1}{n}\right)^n\leq \frac{20}{7}$ by combining the AM-GM inequality with a creative telescoping idea.

Another possible approach, once it is proved that the sequence $\left(1+\frac{1}{n}\right)^n$ is increasing and convergent to $e$, it is to notice that $x(1-x)$ is non-negative and bounded by $\frac{1}{4}$ on $(0,1)$, hence

$$ 0\leq -1+\frac{3}{e} = \int_{0}^{1}x(1-x) e^{-x}\,dx \leq \frac{1}{4} $$ immediately leads to $e<3$.

The layman's way is just to recall that $e^{-x}$ is an entire function, hence $\frac{1}{e}$ is represented by a fast-convergent series: $$ \frac{1}{e} = \frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\ldots $$ and due to the alternating signs: $$ \frac{1}{3}=\frac{1}{2!}-\frac{1}{3!}<\frac{1}{e}< \frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!} = \frac{3}{8}$$ hence $e\in\left(\tfrac{8}{3},3\right).$

I am not sure if this will work for $\frac{2}{3}$ but something similar does for $\frac{1}{2}$ ...

For $k \geq 2$ we have $ k! \geq 2^{k-1}$ \begin{eqnarray*} \sum_{k=0}^{\infty} \frac{1}{k!} = 1+1+ \sum_{k=2}^{\infty} \frac{1}{k!} < 2+\sum_{k=2}^{\infty} \left(\frac{1}{2}\right)^{k-1} =2+ \frac{\frac{1}{2}} {1-\frac{1}{2}} =3. \end{eqnarray*}