In order to study the convergence of $(1+\frac{1}{n})^n$ to $e$,consider the sequences

\begin{equation}

a_n=(1+\frac{1}{n})^n~~~\mbox{and}~~~b_n=(1+\frac{1}{n})^{n+1}

\end{equation}show that

\begin{equation}

a_1<a_{2}<\cdots<e<\cdots<b_3<b_2<b_1

\end{equation}

and that $b_n-a_n\leq \frac{4}{n}$.

 

Proof:According to,$a_1<a_2<\cdots<e$.Then we prove that

\begin{equation}

(1+\frac{1}{n})^{n+1}>(1+\frac{1}{n+1})^{n+2}

\end{equation}

This is simple,because

\begin{equation}

\label{eq:ksdas}

(1+\frac{1}{n+1})^{n+2}=(1+\frac{1}{n+1})^{n+1}(1+\frac{1}{n+1})<(1+\frac{1}{n})^n(1+\frac{1}{n})=(1+\frac{1}{n})^{n+1}

\end{equation}

Then we prove that

\begin{equation}

(1+\frac{1}{n})^{n+1}-(1+\frac{1}{n})^n\leq \frac{4}{n}

\end{equation}

We only need to prove that

\begin{equation}

(1+\frac{1}{n})^n\leq 4

\end{equation}

This is simple,because

\begin{equation}

e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots<3

\end{equation}