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.
 A: 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 $e\in\left(\tfrac{8}{3},3\right).$
A: 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*}
A: To answer your question directly $k! = k(k - 1)...1.$ There are $k$ terms in there: $k, k - 1, k - 2, ..., 1.$ If $k > 2$ Then $(k-2)...(2) > (3/2)^{k - 2}$ because each term is greater than $3/2.$ So we just have to see if $k > (3/2)^2 = 9/4.$ Clearly, because $k$ is an integer and $k > 2$ we have that $k > 9/4.$ 
A: As Thomas Andrews pointed out in his comment, you need to show
$1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\cdots < 1+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$
Lemma: For all $n \ge 1$,
$\tag 1 \;2^n \le (n+1)!$
Proof
Let $I_n = \{1,2,3,\dots,n\}$, and consider the symmetric group $S_{n+1}$ on $I_n \cup \{\hat x\}$. We claim that for any subset $J$ of $I_n$ we can associate a permutation $\sigma_J$ in $S_{n+1}$, and that the association is injective,
$\tag 2 \sigma_J = \sigma_K \Rightarrow J = K$
The empty set gets mapped to the identity permutation.
Any singleton, say $\{k\}$, would get mapped to the transposition $(k,\hat x)$.
Any other subset would get mapped to the natural cyclic permuation in $S_n$ that leaves $\hat x$ invariant.
(Example: $\{1,2,4,7\}$ in $I_{12}$ would map to $(1 \; 2\; 4\; 7)$ in $S_{12} \subset S_{13}$)
It is easy to see that this is injective, proving the claim.
The lemma has been proved since the number of subsets of $I_n$ is $2^n$ and the number of permutations in $S_{n+1}$ is $(n+1)!$.
You can check that the inequality (1) is strict when $n = 2$.

You can also verify (1) by inspection. The LHS has $n$ factors all equal to $2$, while the RHS also has $n$ factors all greater than or equal to $2$.

Another upper bound using elementary techniques:
$e \lt 65/24 + 2^{-5} = 2.739583333$
The inspection method from above can be improved to show that
$\tag 3 \text{For all } n \ge 6 \text{, } \; 2^n \le (n-1)!$
So,
$1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}+\cdots \lt$
$\quad (1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24})+2^{-6}+2^{-7}+\cdots =$
$\quad (65/24)+2^{-5}(2^{-1}+2^{-2}+\cdots) = 65/24 + 2^{-5} $
You can continue this to get upper bounds arbitrarily close to $e$, the next step using
$\tag 4\text{For all } n \ge 8 \text{, } \; 2^n \le (n-2)!$
