Question

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

Short answer

The p-series \(\sum_{n=1}^{\infty} \frac{1}{n^{3}}\) converges.

Step-by-step solution

Set up the Integral

According to the Integral Test, we compare the series to the integral \( \int_{1}^{\infty}\frac{1}{x^{3}}dx\)

Calculate the Integral

We can calculate this integral as follows: \( \int_{1}^{\infty}\frac{1}{x^{3}}dx = [-\frac{1}{2x^{2}}]|_{1}^{\infty} \).

Evaluate the Integral

First, plug in \( \infty \) and then subtract the result when \( 1 \) is plugged in: \( \lim_{t\to \infty}[-\frac{1}{2t^{2}}] + \frac{1}{2} \). Because as \( t \) approaches \( \infty \), \( \frac{1}{2t^{2}} \) approaches \( 0 \), this evaluates to \( 0 + \frac{1}{2} \).

Determine Convergence or Divergence

Since the integral is finite, it converges. Therefore, according to the Integral Test, the series also converges.