# Does a_n=n*{(3/n)^(1/n)} converge? If so what is the limit?

Dec 2, 2016

${\lim}_{n \to \infty} {a}_{n} = \infty$, thus ${a}_{n}$ diverges.

#### Explanation:

${\lim}_{n \to \infty} {a}_{n} = {\lim}_{n \to \infty} n {\left(\frac{3}{n}\right)}^{\frac{1}{n}}$

$= {\lim}_{n \to \infty} {e}^{\ln} \left(n {\left(\frac{3}{n}\right)}^{\frac{1}{n}}\right)$

$= {\lim}_{n \to \infty} {e}^{\ln \left(n\right) + \ln \left({\left(\frac{3}{n}\right)}^{\frac{1}{n}}\right)}$

$= {\lim}_{n \to \infty} {e}^{\ln \left(n\right) + \frac{1}{n} \ln \left(\frac{3}{n}\right)}$

$= {e}^{{\lim}_{n \to \infty} \ln \left(n\right) + \ln \frac{\frac{3}{n}}{n}}$

Where the final step follows from the continuity of $f \left(x\right) = {e}^{x}$. Working on the new limit...

${\lim}_{n \to \infty} \ln \left(n\right) + \ln \frac{\frac{3}{n}}{n} = {\lim}_{n \to \infty} \ln \left(n\right) + \frac{\ln \left(3\right) - \ln \left(n\right)}{n}$

$= {\lim}_{n \to \infty} \ln \frac{3}{n} + \ln \left(n\right) \left(1 - \frac{1}{n}\right)$

As $n \to \infty$, we have

• $\ln \frac{3}{n} \to 0$
• $\ln \left(n\right) \to \infty$
• $1 - \frac{1}{n} \to 1$

$\implies {\lim}_{n \to \infty} \ln \frac{3}{n} + \ln \left(n\right) \left(1 - \frac{1}{n}\right) = 0 + \infty \cdot 1 = \infty$

Substituting this back in, we get

${\lim}_{n \to \infty} {a}_{n} = {e}^{{\lim}_{n \to \infty} \ln \left(n\right) + \ln \frac{\frac{3}{n}}{n}}$

$= {e}^{\infty}$

$= \infty$

Thus ${a}_{n}$ diverges as $n \to \infty$