Question

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

Answer 1

Let

`a_n={9^n}/{n}`. `Rightarrow a_{n+1}={9^{n+1}}/{n+1}`

By Ratio Test,

`lim_{n to infty}|{a_{n+1}}/{a_n}|=lim_{n to infty}|{{9^{n+1}}/{n+1}}/{{9^n}/{n}}|`

by cancelling out common factors,

`=lim_{n to infty}|{9n}/{n+1}|`

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

`=9lim_{n to infty}|{1}/{1+1/n}|=9cdot 1/{1+0}=9 ge 1`

Hence, the series diverges.

I hope that this was helpful.