# Is the sequence a_n=(1+3/n)^(4n) convergent or divergent?

## My issue is that the exponent is 4n, not n... I only know how to solve if it were just $n$

Apr 23, 2018

$\text{See explanation}$

#### Explanation:

${a}_{n} = {\left({\left(1 + \frac{3}{n}\right)}^{4}\right)}^{n}$
$= {\left({\left({\left(1 + \frac{3}{n}\right)}^{2}\right)}^{2}\right)}^{n}$
$= {\left({\left(1 + \frac{6}{n} + \frac{9}{n} ^ 2\right)}^{2}\right)}^{n}$
$= {\left(1 + \frac{36}{n} ^ 2 + \frac{81}{n} ^ 4 + \frac{12}{n} + \frac{18}{n} ^ 2 + \frac{108}{n} ^ 3\right)}^{n}$
$= {\left(1 + \frac{12}{n} + \frac{54}{n} ^ 2 + \frac{108}{n} ^ 3 + \frac{81}{n} ^ 4\right)}^{n}$

$\text{Note that you could more easily apply Euler's limit here :}$

${\lim}_{n \to \infty} {\left(1 + \frac{1}{n}\right)}^{n} = e = 2.7182818 \ldots .$

$\implies {\lim}_{n \to \infty} {\left(1 + \frac{3}{n}\right)}^{12 \cdot \frac{n}{3}} = {e}^{12} = 162754.79 \ldots .$

$\text{So the sequence grows very big but not infinitely big, so it}$
$\text{converges.}$