<s>[INST] Prove that $a_n = \{\left(1+\frac{1}{n}\right)^n\}$ is bounded sequence, $ n\in\mathbb{N}$ How to prove the following:
$a_n = \left\{\left(1+\frac{1}{n}\right)^n\right\}$ is bounded sequence, $ n\in\mathbb{N}$
 [/INST] There was a comment that you can use $\text{AM} \ge \text{GM}$ to prove the boundedness of this.
Here is a proof. As a side effect, we also prove the convergence.
First we show that $x_n = \left(1 + \frac{1}{n}\right)^n$ is monotonically increasing. We prove this, using $\text{AM} \ge \text{GM}$.
We have that, by taking $n$ copies of $\left(1 + \frac{1}{n}\right)$ and one copy of $1$ that,
$$\frac{\left(1 + \frac{1}{n}\right) + \dots + \left(1 + \frac{1}{n}\right) + 1}{n+1} \gt \sqrt[n+1]{\left(1 + \frac{1}{n}\right)^n}$$
i.e.
$$ \frac{n+2}{n+1} \gt \sqrt[n+1]{\left(1 + \frac{1}{n}\right)^n}$$
and so
$$\left(1 + \frac{1}{n+1}\right)^{n+1} \gt \left(1 + \frac{1}{n}\right)^n$$
Now we use $\text{AM} \ge \text{GM}$ (again!). 
We take $n$ copies of $1$ and one copy of $\frac{1}{2}$ to get
$$\frac{n + \frac{1}{2}}{n+1} \ge \sqrt[n+1]{\frac{1}{2}}$$
i.e.
$$ 2^{\frac{1}{n+1}} \ge \frac{2n+2}{2n+1}$$
i.e
$$ 2 \ge \left(1 + \frac{1}{2n+1}\right)^{n+1} $$
And so
$$ 4 \ge \left(1 + \frac{1}{2n+1}\right)^{2n+1} $$
Since the sequence is monotonically increasing, this bound applies to the whole sequence.
(Side effect: Since the sequence is monotonic, and bounded, it is convergent)
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