# Calc6.68

Discuss the convergence or divergence of the series $\sum _{{n=1}}^{{\infty }}{\frac {n}{3^{n}}}$.

$\lim _{{n\rightarrow \infty }}{\sqrt[ {n}]{{\frac {n}{3^{n}}}}}=\lim _{{n\rightarrow \infty }}{\frac {{\sqrt[ {n}]{n}}}{3}}={\frac {1}{3}}$

Since the limit is less than 1, by the root test, this series converges absolutely. It is true that

$\lim _{{n\rightarrow \infty }}{\sqrt[ {n}]{n}}=1$,

which we just used for this problem. I will show this is true.

Let $y=\lim _{{n\rightarrow \infty }}{\sqrt[ {n}]{n}}$. From here, we can take the natural log of both sides, and in fact, on the right side, we can put the natural log on the inside of the limit. This is the trick for finding this limit analytically. So, we have

$\ln y=\lim _{{n\rightarrow \infty }}\ln {\sqrt[ {n}]{n}}=\lim _{{n\rightarrow \infty }}{\frac {1}{n}}\ln n=\lim _{{n\rightarrow \infty }}{\frac {\ln n}{n}}$

Here, plugging in infinity gives an indeterminate form so we can use L'Hopitals rule, provided the limit exists. This gives us

$\ln y=\lim _{{n\rightarrow \infty }}{\frac {\ln n}{n}}=\lim _{{n\rightarrow \infty }}{\frac {{\frac {1}{n}}}{1}}=\lim _{{n\rightarrow \infty }}{\frac {1}{n}}=0$

So, we have $\ln y=0$. Thus, $y=1$ as claimed. Thus