Question

Use the comparison test to determine whether the following series converge. $$ \sum_{n=1}^{\infty} \frac{\sin ^{2} n}{n^{2}} $$

Short answer

The series \( \sum_{n=1}^{\infty} \frac{\sin^2 n}{n^2} \) converges.

Step-by-step solution

Identify the series

Given the series \( \sum_{n=1}^{\infty} \frac{\sin^2 n}{n^2} \). We need to determine its convergence using the comparison test.

Choose a comparable series

Recall that the sine function is bounded: \( 0 \leq \sin^2 n \leq 1 \). So, \( \frac{\sin^2 n}{n^2} \leq \frac{1}{n^2} \). We will compare it to the known convergent series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \), which is a p-series with \( p = 2 > 1 \).

Apply the Comparison Test

According to the comparison test, if \( 0 \leq a_n \leq b_n \) for all \( n \), and \( \sum b_n \) converges, then \( \sum a_n \) converges as well. Here, \( a_n = \frac{\sin^2 n}{n^2} \) and \( b_n = \frac{1}{n^2} \). Since \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) converges and \( \frac{\sin^2 n}{n^2} \leq \frac{1}{n^2} \), the original series \( \sum_{n=1}^{\infty} \frac{\sin^2 n}{n^2} \) converges.

Key concepts

Convergent Series

When we say a series is convergent, we mean that as we add more and more terms of the series, the sum gets closer to a certain number, or limit. This limit is a fixed value, which means it doesn't keep changing. A series \( \sum_{n=1}^{\infty} a_n \) is convergent if the sequence of its partial sums \( S_N = a_1 + a_2 + \ldots + a_N \) approaches a specific value as \( N \rightarrow \infty \).Convergent series are important in calculus and analysis because they describe situations where adding up an infinite number of terms results in a finite sum. The comparison test is a useful tool for determining convergence. By comparing the series in question to another series that we know converges, we can sometimes determine the behavior of the unknown series. In our example, we used the comparison test to establish that the series \( \sum_{n=1}^{\infty} \frac{\sin^2 n}{n^2} \) converges because it was compared to the convergent p-series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \).

P-Series

A p-series is a type of mathematical series of the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p \) is a constant. A key characteristic of p-series is their convergence or divergence based on the value of \( p \):

• If \( p > 1 \), the series converges.
• If \( p \leq 1 \), the series diverges.

This rule makes p-series useful for comparison tests, as they provide a readily known reference point on which to base our judgments about the behavior of other series. In the series \( \sum_{n=1}^{\infty} \frac{\sin^2 n}{n^2} \), the series we compared it to was \( \sum_{n=1}^{\infty} \frac{1}{n^2} \), which is a p-series with \( p = 2 \). Since \( 2 > 1 \), we know this p-series is convergent, allowing us to apply the comparison test effectively.

Bounded Function

A function is said to be bounded if there is a limit to how large or small the function's values can be. Specifically, for a function \( f(x) \), it is bounded if there are real numbers \( M \) and \( m \) such that \( m \leq f(x) \leq M \) for all values in the domain.In our example, we used this idea when considering \( \sin^2 n \). The sine function, \( \sin n \), is naturally bounded because its values always lie between -1 and 1. Therefore, \( \sin^2 n \) ranges between 0 and 1 because squaring any real number gets rid of negative values:

• \( 0 \leq \sin^2 n \leq 1 \).

This bounded nature allowed us to simplify the series to \( \frac{\sin^2 n}{n^2} \leq \frac{1}{n^2} \), helping us use the comparison test effectively, as we compared it to the simpler p-series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \). Knowing that \( \sin^2 n \) is bounded made it easier to draw conclusions about the original series.