Question

Use the test of your choice to determine whether the following series converge. $$\sum_{k=1}^{\infty} \sin ^{2} \frac{1}{k}$$

Short answer

Answer: The given series converges.

Step-by-step solution

Prove the inequality

We want to prove that $$0 \le \sin^2\frac{1}{k} \le \frac{1}{k^2}$$ for all k >= 1. Since $$0 ≤ \sin(\frac{1}{k}) ≤ \frac{1}{k}$$, squaring both sides gives: $$0 ≤ \sin^2(\frac{1}{k}) ≤ \frac{1}{k^2}$$ So, the inequality is proved.

Compare with the harmonic series

We will now compare the original series with the harmonic series: $$\sum_{k=1}^{\infty} \sin^{2} \frac{1}{k}$$ and $$\sum_{k=1}^{\infty} \frac{1}{k^2}$$ We know that: $$0 \le \sum_{k=1}^{\infty} \sin^{2} \frac{1}{k} \le \sum_{k=1}^{\infty} \frac{1}{k^2}$$ We also know that the harmonic series of $$\frac{1}{k^2}$$ converges (this is the p-series with p=2, which converges when p>1). Therefore, by the Direct Comparison Test, the given series also converges: $$\sum_{k=1}^{\infty} \sin^{2} \frac{1}{k}$$ converges.

Key concepts

Direct Comparison Test

Understanding the Direct Comparison Test (DCT) is essential in determining whether a series converges or diverges. The test involves comparing two series term-by-term. Here's the core principle:
If we have two series, \( \sum a_k \) and \( \sum b_k \), such that \( 0 \leq a_k \leq b_k \) for all \( k \) in their domain, and if \( \sum b_k \) converges, then so does \( \sum a_k \). Conversely, if \( \sum a_k \) diverges, then \( \sum b_k \) must also diverge.
This test is particularly helpful when dealing with series that are not immediately recognizable as convergent or divergent. By comparing a series with a known counterpart, we can often draw conclusions about its behavior.

Harmonic Series

The harmonic series is the infinite series \( \sum_{k=1}^{\infty} \frac{1}{k} \). Despite its simple form, this series diverges, meaning its terms do not add up to a finite value. This might be counterintuitive because the individual terms of the series become very small as \( k \) increases. However, their sum grows without bound, a fact that can be proven through various tests for divergence.
Intuitively, the divergence of the harmonic series can be understood by grouping terms. As more and more terms are added, their collective sum increases, albeit more slowly, but it never stops growing. The harmonic series is a classic example in calculus to demonstrate how a series with diminishing terms can still fail to converge.

p-Series

p-Series are a type of series with the general form \( \sum_{k=1}^{\infty} \frac{1}{k^p} \) for \( p > 0 \). The convergence of a p-series depends on the value of \( p \). For \( p \leq 1 \) the series diverges, and for \( p > 1 \) it converges. This fact is critical in understanding the behavior of series and in comparison tests.
For instance, a series with \( p = 2 \) is known to converge, and this property is useful when using the Direct Comparison Test. In our exercise, comparing the given series to a p-series with \( p = 2 \) helped us conclude the given series' convergence.

Inequalities in Calculus

Inequalities play a vital role in calculus, especially in analyzing series and functions. Establishing inequalities can help us prove convergence, find bounds, and understand the behavior of functions. As shown in the exercise, by proving the inequality \( 0 \leq \sin^2(\frac{1}{k}) \leq \frac{1}{k^2} \), we can make a comparison to a known convergent series.
Ensuring that the inequalities are valid for all terms in the series is crucial. The squaring of both sides of \( 0 \leq \sin(\frac{1}{k}) \leq \frac{1}{k} \) is legitimate because both sides are non-negative. Care should be taken with inequalities, particularly when dealing with absolute values or functions that can take on negative values, as squaring can affect the inequality's direction.