Question

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

Short answer

Answer: The series converges.

Step-by-step solution

Set up the Ratio for Limit Comparison Test

Compute the limit as \(k\) goes to infinity of the ratio \(\frac{\sin ^{2}\left(\frac{1}{k}\right)}{k^{-2}}\).

Compute the Limit

Using L'Hopital's rule and the fact that \(\lim_{x\to 0} \frac{\sin x}{x} = 1\), we compute the following limit: $$\lim_{k \to \infty} \frac{\sin ^{2}\left(\frac{1}{k}\right)}{k^{-2}} = \lim_{k\to\infty}\frac{\sin^2\left(\frac{1}{k}\right)}{\frac{1}{k^2}} = \lim_{u\to 0}\frac{\sin^2(u)}{u^2}$$ Now, rewrite \(\sin^2(u)\) as \((\sin(u))^2\). Then, the limit is: $$\lim_{u \to 0} \frac{(\sin(u))^2}{u^2} = \left(\lim_{u\to 0}\frac{\sin(u)}{u}\right)^2$$ Since \(\lim_{x\to 0} \frac{\sin x}{x} = 1\), we have: $$\left(\lim_{u\to 0}\frac{\sin(u)}{u}\right)^2 = (1)^2 = 1$$

Determine Convergence

Since the limit is \(1\), which is a positive finite value, by the Limit Comparison Test, we can conclude that the series \(\sum_{k=1}^{\infty} \sin ^{2}\left(\frac{1}{k}\right)\) has the same convergence behavior as the series \(\sum_{k=1}^{\infty} k^{-2}\). Since the latter converges, the series \(\sum_{k=1}^{\infty} \sin ^{2}\left(\frac{1}{k}\right)\) also converges.

Key concepts

Limit Comparison Test

The Limit Comparison Test is a useful tool for determining the convergence of an infinite series. Suppose you have two series, say \( \sum a_k \) and \( \sum b_k \). The test says that if you compute the limit \( \lim_{k \to \infty} \frac{a_k}{b_k} = L \), where \( L \) is a positive, finite number, then both series will either converge or diverge together. In this specific exercise, the function \( a_k = \sin^2\left(\frac{1}{k}\right) \) is compared with \( b_k = k^{-2} \). The limit \( \lim_{k \to \infty} \frac{\sin^2(\frac{1}{k})}{k^{-2}} \) simplifies to \( 1 \), a finite number, implying both series share the same behavior. Since \( \sum_{k=1}^{\infty} k^{-2} \) is known to converge, the original series \( \sum_{k=1}^{\infty} \sin^2\left( \frac{1}{k} \right) \) also converges.

L'Hopital's Rule

L'Hopital's rule is an amazing tool for solving tricky limits that come in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). The rule states that if the limit \( \lim_{x\to c} \frac{f(x)}{g(x)} \) yields an indeterminate form, you can compute \( \lim_{x\to c} \frac{f'(x)}{g'(x)} \) instead, provided the derivatives exist and the new limit is defined. In this problem, we used L'Hopital's rule to resolve the indeterminate form in the expression \( \lim_{u\to0} \frac{(\sin u)^2}{u^2} \). By applying this rule twice, or realizing that \( \lim_{u\to0} \frac{\sin u}{u} = 1 \), we find \( (\sin u)^2 \sim u^2 \). Therefore, the limit simplifies to 1, helping us validate the convergence using the Limit Comparison Test.

Convergent Series

A convergent series is essentially a series whose terms approach a finite sum as more and more terms are added. Mathematically, an infinite series \( \sum_{k=1}^{\infty} a_k \) is convergent if the sequence of partial sums \( S_n = a_1 + a_2 + \ldots + a_n \) approaches a specific number as \( n \to \infty \). In the given exercise, once we established that \( \sum_{k=1}^{\infty} \sin^2\left(\frac{1}{k}\right) \) is of the same type as \( \sum_{k=1}^{\infty} k^{-2} \) through the Limit Comparison Test, we concluded it converges since \( \sum_{k=1}^{\infty} k^{-2} \) is a well-known convergent series called the p-series with \( p = 2 \).

Infinite Series

An infinite series is a sum of an infinite sequence of terms, and it's a fascinating concept in mathematics. Formally, it involves adding up infinitely many numbers, one after the other. Such sums can be tricky because they don't behave like our typical finite sums. They can converge (add up to a finite number) or diverge (not add up to any specific number). The series \( \sum_{k=1}^{\infty} \sin^2\left(\frac{1}{k}\right) \) from the exercise is an infinite series. By comparing it to other well-known series such as \( \sum_{k=1}^{\infty} k^{-2} \) using the Limit Comparison Test, we determined its behavior. This shows how a seemingly infinite process can still have meaningful results in mathematics.