<s>[INST] Prove $(1 + \frac{1}{n})^n$ is bounded above I've checked similar questions on the site but couldn't find satisfactory solutions or hints. 
Also, is there a more general approach to proving whether a given sequence is bounded below or above?
 [/INST] By the binomial formula:
\begin{eqnarray*}
\left(1+\frac{1}{n}\right)^n=1+1+\sum_{k=2}^n\binom{n}{k}\cdot\left(\frac{1}{n}\right)^k.
\end{eqnarray*}
Notice that
\begin{eqnarray*}
\binom{n}{k}\cdot\left(\frac{1}{n}\right)^k=\frac{n(n-1)\cdots(n-k+1)}{n^k}\cdot\frac{1}{k!}<\frac{1}{k(k-1)}=\frac{1}{k-1}-\frac{1}{k}.
\end{eqnarray*}
So we get
\begin{eqnarray*}
\left(1+\frac{1}{n}\right)^n<2+\sum_{k=2}^n\left(\frac{1}{k-1}-\frac{1}{k}\right)=3-\frac{1}{n}<3.
\end{eqnarray*}
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