How do you use the Ratio Test on the series sum_(n=1)^oo9^n/nn=19nn ?

1 Answer
Oct 5, 2014

Let

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