<s>[INST] Proving this inequality $\left(1+\frac{1}{n}\right)^n < 3$ How would I prove this, particularly using a method using a geometric series and binomial coefficient.
I'm having trouble giving a reason for $(\star)$ especially.
This is what I did  
For $n=1$, LHS = $(1+1)^1 = 2 < 3$ = RHS.
Thus, for $n=2$,
$$\left(1+\frac{1}{n}\right)^n \equiv {\large{\sum_{k=0}^n}} \begin{pmatrix}n \\ k \end{pmatrix}\frac{1}{n^k} \\ = {\large{\sum_{k=0}^n}}\frac{n!}{k!(n-k)!n^k} \\ < \sum_{k=0}^n \frac{1}{k!} \\ < \sum_{k=0}^\infty \frac{1}{k!} \\ < \sum_{k=0}^\infty \frac{1}{(\frac{3}{2})^k} \qquad (\star)\\ = 3. $$  
$(\star)$ since $k>2 > \frac{3}{2} \implies k! > 2 > \frac{3}{2}$ and $k! > (\frac{3}{2})^k$ for $k>2$.  
I just know this to be true (the last step of raising 1.5 to the power $k$) by checking my calculator, but I don't see a reason why the factorial grows faster or becomes larger than the exponential.
 [/INST] 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 $e\in\left(\tfrac{8}{3},3\right).$
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