How do you use the Ratio Test on the series sum_(n=1)^oo9^n/n ?

Oct 5, 2014

Let

${a}_{n} = \frac{{9}^{n}}{n}$. $R i g h t a r r o w {a}_{n + 1} = \frac{{9}^{n + 1}}{n + 1}$

By Ratio Test,

${\lim}_{n \to \infty} | \frac{{a}_{n + 1}}{{a}_{n}} | = {\lim}_{n \to \infty} | \frac{\frac{{9}^{n + 1}}{n + 1}}{\frac{{9}^{n}}{n}} |$

by cancelling out common factors,

$= {\lim}_{n \to \infty} | \frac{9 n}{n + 1} |$

by pulling 9 out of the limit and dividing the numerator and denominator by $n$,

$= 9 {\lim}_{n \to \infty} | \frac{1}{1 + \frac{1}{n}} | = 9 \cdot \frac{1}{1 + 0} = 9 \ge 1$

Hence, the series diverges.

I hope that this was helpful.