Question

How do you use the integral test to determine the convergence or divergence of `Sigma 1/n^(1/3)` from `[1,oo)`?

Answer 1

The sum `sum_(n=1)^oo1/n^(1/3)` diverges.

Explanation:

To test the convergence/divergence of `sum_(n=1)^oo1/n^(1/3)` we examine the integral `int_1^oo1/x^(1/3)dx` which can be rewritten as `int_1^oox^(-1/3)dx`. Using the antiderivative method of integration, we find this to be `3/2(oo)^(2/3)-3/2(1)^(2/3)=oo`. Since the value of the integral goes to infinity, the sum `sum_(n=1)^oo1/n^(1/3)` diverges by the integral test.