Question

Use the Integral Test to determine the convergence or divergence of the \(p\) -series. $$ \sum_{n=1}^{\infty} \frac{1}{n^{1 / 3}} $$

Short answer

The infinite p-series \( \sum_{n=1}^{\infty} 1/n^{1 / 3}\) diverges.

Step-by-step solution

Identify the series type and function

Identify the function in the given series. The series is a p-series and the function is \( f(n)= 1/n^{1 / 3}\).

Demonstrate criteria to apply Integral Test

Show that the function f(n)= 1/n^{1 / 3} is continuous, positive, and decreasing on [1, ∞). Since \(n^{1 / 3}\) is a cube root function, it is continuous and positive for \(n \geq 1\), and it's decreasing as the denominator increases: the larger the \(n\), the smaller the \(f(n)\). Therefore, the conditions to apply the Integral Test are met.

Apply the Integral Test

Take the integral from 1 to ∞ of the function \(f(x)= 1/x^{1 / 3}\) to determine if it converges or diverges. The integral is \(\int_{1}^{\infty} x^{-1 / 3} dx\). Evaluate the integral to see if it converges or diverges. The antiderivative is \(3/2*x^{2 / 3}\), and evaluating from 1 to ∞ gives \(\infty\) - \(3/2\) which is \(\infty\). Therefore, the integral diverges.