How is $a_n=(1+1/n)^n$ monotonically increasing and bounded by $3$? I was reading about how completeness is required for limits. And I came across this:

the sequence $a_n=(1+1/n)^n$ is monotonically increasing and bounded by 3 and so we expect it to converge, but that it does not converge within $\mathbb{Q}$. More generally it stands to reason that any sequence of real numbers which is increasing and bounded must converge to some real number. This is a consequence of completeness of $IR$

My question is: How is the mentioned sequence monotonically increasing and bounded by $3$ ? 
 A: The monotonicity follows from the AM-GM inequality for $n+1$ points. Taking $x_1 = x$ and $x_2 = \cdots = x_{n+1} = y$, we get
$$
\sqrt[n+1]{xy^n} \leq \frac{x+ny}{n+1}.
$$
In particular, taking $x = 1$ and $y = 1+\frac{1}{n}$ yields $a_n \leq a_{n+1}$ ($n \geq 1$).
Now as another special case, take $x = 1$ and $y = 1-\frac{1}{n}$ to get
$$
b_n \geq b_{n+1} \qquad (n \geq 1),
$$
where $b_n := \left( 1-\frac{1}{n}\right)^{-n}$. We let you verify that
$$
b_n = a_{n-1} \left( 1+\frac{1}{n-1}\right) \geq a_{n-1}, \qquad (n \geq 2).
$$
Hence
$$
a_1 \leq a_2 \leq a_3 \leq \cdots \leq b_3 \leq b_2 \leq b_1
$$
and we deduce that $a_n \leq b_m$ for all $n, m \in \mathbb{N}$. In particular, taking $m = 6$, we get
$$
a_n \leq \left( \frac{6}{5} \right)^6 \leq 3.
$$
A: From Rudin's PMA Theorem $3.31$, by the Binomial theorem,
$$ t_n=\left( 1 + \frac{1}{n} \right)^n $$
$$= 1 + 1 + \underbrace{\frac{1}{2!}\left(1-\frac{1}{n}\right)}_{\text{term } 2} + \underbrace{\frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)}_{\text{term } 3} + \ldots + \frac{1}{n!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{n-1}{n}\right). $$
Since each bracket increases as $n$ increases, each term increases as $n$ increases also, because products and sums of increasing positive functions is also increasing.
Therefore, for all $n\geq 2,$ and all $2\leq k \leq n,$ the $k$-th term of $t_{n+1}$ is greater than the $k$-th term of $t_n.$
So we see that $\left( 1 + \frac{1}{n} \right)^n$ is increasing.
Furthermore, the Binomial expansion above is
$$ \leq 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \ldots + \frac{1}{n!} = 1 + 1 + \frac{1}{1\cdot 2} + \frac{1}{1\cdot 2\cdot 3} + \frac{1}{1\cdot 2\cdots n} $$
$$ < 1+1+ \frac{1}{2} + \frac{1}{2^2} + \ldots + \frac{1}{2^{n-1}} <3. $$
