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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.

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.

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Answer 2

Elijah Abram

To use the integral test to determine the convergence or divergence of the series Σ 1/n^(1/3) from n = 1 to infinity:

Evaluate the integral ∫(1/x^(1/3)) dx from 1 to infinity.
If the integral converges, then the series Σ 1/n^(1/3) also converges. If the integral diverges, then the series Σ 1/n^(1/3) also diverges.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.