How do you apply the ratio test to determine if Sigma 1/n^3 from n=[1,oo) is convergent to divergent?

1 Answer
Mar 21, 2018

It's convergent. You may use the integral test to prove this.

Explanation:

The ratio test isn't the correct test. I would use the integral test.

We need to evaluate the integral

=int_1^oo 1/n^3 dn

=lim_(t-> oo) int _1^t 1/n^3 dn

=lim_(t-> oo) [-1/2n^-2]_1^t

=lim_(t->oo) -1/2t^-2 - (-1/2(1)^-2)

=0 + 1/2

= 1/2

Since int_1^oo 1/n^3 dn is convergent, than the series is also convergent.

Hopefully this helps!