<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] As Weson Jiang notes, $a_n = \left(n + \frac{1}{n}\right)^n > n^n$, and $n^n\to\infty$ as $n\to\infty$, so $a_n\to\infty$ as well.
You might have wanted to show that $\left\{b_n\right\}_{n\in\Bbb{N}} = \left\{\left(1 + \frac{1}{n}\right)^n\right\}_{n\in\Bbb{N}}$ is a bounded sequence. The binomial theorem implies $\left(1 + \frac{1}{n}\right)^n = \sum_{k = 0}^n\begin{pmatrix}n\\k\end{pmatrix}\frac{1}{n^k}$, so
\begin{align*}
\left(1 + \frac{1}{n}\right)^n &= \sum_{k = 0}^n\begin{pmatrix}n\\k\end{pmatrix}\frac{1}{n^k}\\
&=\frac{1}{1} + n\frac{1}{n} + \begin{pmatrix}n\\2\end{pmatrix}\frac{1}{n^2} + \ldots + \frac{1}{n^n}\\
&\leq 1 + 1 + \frac{n^2}{2!}\frac{1}{n^2} + \frac{n^3}{3!}\frac{1}{n^3} + \ldots + \frac{n^n}{n!}\frac{1}{n^n}\\
&< 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \ldots + \frac{1}{n!} + \ldots\\
&< 1 + 1 + \frac{1}{2} + \frac{1}{2^2} + \ldots + \frac{1}{2^n} + \ldots\\
&= 1 + \sum_{k = 0}^{\infty}\frac{1}{2^k}\\
&= 1 + 2 = 3,
\end{align*}
so the sequence is bounded above by $3$.
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