<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] 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*}
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