# How do you determine whether sum n/3^(n+1) from 1 to infinity converges or diverges?

May 2, 2015

The Ratio Test will show it converges.

The Ratio Test says:
The series ${\sum}_{n = 1}^{\setminus \infty} {a}_{n}$ converges if $\setminus {\lim}_{n \setminus \to \setminus \infty} \setminus \left\mid \setminus \right\mid \frac{{a}_{n + 1}}{{a}_{n}} < 1$, diverges if $\setminus {\lim}_{n \setminus \to \setminus \infty} \setminus \left\mid \setminus \right\mid \frac{{a}_{n + 1}}{{a}_{n}} > 1$, and is inconclusive if $\setminus {\lim}_{n \setminus \to \setminus \infty} \setminus \left\mid \setminus \right\mid \frac{{a}_{n + 1}}{{a}_{n}} = 1$.

For your specific problem, ${a}_{n} = \setminus \frac{n}{{3}^{n + 1}}$ and ${a}_{n + 1} = \setminus \frac{n + 1}{{3}^{n + 2}}$ so

$\setminus {\lim}_{n \setminus \to \setminus \infty} \setminus \left\mid \setminus \right\mid \frac{{a}_{n + 1}}{{a}_{n}} = \setminus {\lim}_{n \setminus \to \setminus \infty} \setminus \left\mid \setminus \right\mid \frac{\setminus \frac{n + 1}{{3}^{n + 2}}}{\setminus \frac{n}{{3}^{n + 1}}} =$
$= \setminus {\lim}_{n \setminus \to \setminus \infty} \setminus \left\mid \setminus \frac{n + 1}{{3}^{n + 2}} \setminus \cdot \setminus \frac{{3}^{n + 1}}{n} \right\mid = \setminus {\lim}_{n \setminus \to \setminus \infty} \setminus \left\mid \setminus \frac{n + 1}{3 n} \right\mid =$
$= \setminus \frac{1}{3} \setminus {\lim}_{n \setminus \to \setminus \infty} \setminus \left\mid \setminus \frac{n + 1}{n} \right\mid = \setminus \frac{1}{3} < 1$, so the series converges.