Tenemos las siguientes desigualdades:
$\left(1+\dfrac{1}{n}\right)^n = 1 + 1 + \dfrac{1}{2!}\left(1-\dfrac{1}{n}\right)+\dfrac{1}{3!}\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right) + \ldots \leq 2 + \dfrac{1}{2} + \dfrac{1}{2^2} + \dots =3$
Del mismo modo,
$\begin{align*}
\left(1-\dfrac{1}{n^2}\right)^n &= 1 - {n \choose 1}\frac{1}{n^2} + {n \choose 2}\frac{1}{n^4} + \dots\\
&= 1 - \frac{1}{n} + \dfrac{1}{2!n^2}\left(1-\dfrac{1}{n}\right) - \dfrac{1}{3!n^3}\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right) + \ldots
\end{align*} $
Así, $$ | \left (1-\dfrac {1} {n ^ 2} \right) ^ {n} - \left (1-\frac {1} {n} \right)| \leq \dfrac{1}{2n^2} + \dfrac{1}{2^2n^2} + \dfrac{1}{2^3n^2} + \ldots = \dfrac{1}{n^2}.
$$
Ahora,
$\begin{align*}
\left(1+\frac{1}{n+1}\right)^{n+1} - \left(1+\frac{1}{n}\right)^n &= \left(1+\frac{1}{n}-\frac{1}{n(n+1)}\right)^{n+1}-\left(1+\frac{1}{n}\right)^n \\
&=\left(1+\frac{1}{n}\right)^{n+1}\left\{ \left( 1- \frac{1}{(n+1)^2}\right)^{n+1} - \frac{n}{n+1}\right\}\\
&= \left(1+\frac{1}{n}\right)^{n+1}(1 - \frac{1}{n+1} + \text{O}(\frac{1}{n^2}) - \frac{n}{n+1})\\
& = \left(1+\frac{1}{n}\right)^{n+1}\text{O}(\frac{1}{n^2}) \\
&= \text{O}(\frac{1}{n^2}) \text{ (since %#%#% is bounded) }.
\end{align*} $
Así, dejando que $(1+1/n)^{n+1}$ tenemos, $a_k = (1+1/k)^k$ $|a_{k+1}-a_k| \leq C/k^2$ y por lo tanto, $C$ $\sum_{ k \geq n } | a_{k+1} - a_k | \to 0$.
Desde $n \to \infty$. Así que dado $|a_n - a_m| \leq \sum_{ k \geq \min\{m,n\}} |a_{k+1} - a_k|$eligió $\epsilon > 0$ tal que $N$ y $\sum_{ k \geq N} |a_k - a_{k+1}| < \epsilon$ $|a_n - a_m| < \epsilon $.
