Question

In the following exercises, use an appropriate test to determine whether the series converges. $$ a_{k}=1 / 2^{\sin ^{2} k} $$

Short answer

The series diverges as standard convergence tests fail to demonstrate bounded behavior.

Step-by-step solution

Understand the Series

We are given the series \( \sum_{k=1}^{\infty} \frac{1}{2^{\sin^2 k}} \). Our task is to determine whether this series converges or diverges.

Apply the Limit Comparison Test

To use the Limit Comparison Test, we need to find a comparison series. Let's compare it with the geometric series \( \sum \frac{1}{2^k} \), which is known to converge. Our task is to compare \( \frac{1}{2^{\sin^2 k}} \) and \( \frac{1}{2^k} \).

Analyze \( \sin^2 k \) Behavior

The function \( \sin^2 k \) oscillates between 0 and 1 for positive integers \( k \). Therefore, \( 0 \leq \sin^2 k \leq 1 \). This implies that \( 2^{0} = 1 \leq 2^{\sin^2 k} \leq 2^{1} = 2 \).

Establish Comparison Bounds

Based on Step 3, \(1 \leq 2^{\sin^2 k} \leq 2 \). So, \( \frac{1}{2^{\sin^2 k}} \) has a behavior similar to \( \frac{1}{2} \leq \frac{1}{2^{\sin^2 k}} \leq 1 \). For large \( k \), the series \( \frac{1}{2^k} \) converges since it's a geometric series with a ratio \( r = \frac{1}{2} < 1 \).

Limit Comparison and Conclusion

To apply the Limit Comparison Test, we evaluate \( \lim_{k \to \infty} \frac{\frac{1}{2^{\sin^2 k}}}{\frac{1}{2^k}} = \lim_{k \to \infty} 2^{k-\sin^2 k} \). Since \( \sin^2 k \) remains bounded and \( k \to \infty \), this limit evaluates to infinity, indicating that the Limit Comparison Test cannot be directly applied here due to indeterminate behavior, suggesting comparing directly to elements.

Use the Ratio Test

We will alternatively employ the Ratio Test. Consider \( a_k = \frac{1}{2^{\sin^2 k}} \); then evaluate \( \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| = \lim_{k \to \infty} 2^{\sin^2 k - \sin^2 (k+1)} \). Since \( \sin^2 (k+1) - \sin^2 k \) remains a small bounded value, this ratio does not significantly change.

Conclusion of the Test and Series Divergence

The ratio of consecutive terms does not consistently stay below 1 after evaluation, suggesting that we cannot conclude convergence via these standard tests. However, due to constant fluctuation and bounds not achieving a convergence indicator in tests, the series seems to not grow to a fixed summable value across limitless \( k \), indicating divergence through non-tight bounds.

Key concepts

Limit Comparison Test

The Limit Comparison Test is a handy tool for studying the convergence of series. It works by comparing the given series to another series whose behavior is already known. To use this test effectively, one generally chooses a series with similar structure, like a geometric or harmonic series.

Here's the process in simple steps:

• Identify a series to compare with. Typically, this is a series you know converges or diverges.
• Calculate the limit of the ratio of the terms of the two series as they approach infinity.
• If the limit exists and is a positive finite number, both series will either converge or diverge together.

This test is useful in many situations but has its limitations. For instance, if the limit doesn't exist or isn't positive and finite, the test can't provide a conclusion, as we saw in the given series.

Geometric Series

A geometric series is a series of the form \[ \sum_{k=0}^{ ext{infinity}} ar^k \]where \( a \) is the first term and \( r \) is the common ratio. These series are fascinating because they have a simple convergence rule: they converge if and only if the absolute value of \( r \) is less than one, \(|r| < 1\).

When convergent, the sum of an infinite geometric series is given by:\[ S = \frac{a}{1-r} \]If we look at the comparison series \( \sum \frac{1}{2^k} \) provided in the solution, it's actually a geometric series. The first term \( a = \frac{1}{2} \) and the common ratio \( r = \frac{1}{2} \), hence it converges since \( |\frac{1}{2}| < 1 \).

Understanding geometric series is crucial as they often appear in various forms in calculus problems, serving as benchmarks for other series.

Ratio Test

The Ratio Test determines the convergence of a series by evaluating the limit of the absolute value of the ratio of consecutive terms. It's particularly useful for series with factorials or exponential terms. The test involves these steps:

• Calculate the ratio \( \left| \frac{a_{k+1}}{a_k} \right| \).
• Take the limit as \( k \to \infty \).
• If the limit is less than 1, the series converges absolutely.
• If greater than 1 or is infinite, the series diverges.
• If equal to 1, the test is inconclusive.

In our example, the Ratio Test was used after the Limit Comparison Test indicated incompleteness. It provided insight into the divergence due to the oscillating behavior of \( \sin^2 k \), which prevented the ratios from converging to values indicative of series convergence.

Divergence

Divergence refers to the behavior of a series that doesn't settle into a fixed sum. Basically, if a series fails to converge, it diverges. Divergence can occur for multiple reasons:

• The terms do not approach zero as \( k \to \infty \).
• The series grows without bound.
• Tests for convergence, like the Ratio Test, indicate values outside the range that would suggest convergence.

In our series \( \sum_{k=1}^{\infty} \frac{1}{2^{\sin^2 k}} \), the inability to apply tests conclusively due to fluctuating values led us to infer divergence. This means that as \( k \) increases indefinitely, the series terms do not sum up to a finite number.

Trigonometric Functions

Trigonometric functions like \( \, \sin \, \) are periodic and oscillate in nature, meaning their values cycle through a specific range. For an integer \( k \), \( \sin^2 k \) ranges between 0 and 1. This property was pivotal in the original exercise, as \( \sin^2 k \) influenced the terms \( 2^{\sin^2 k} \).

Understanding trigonometric functions helps in evaluating series where these functions play a role, especially when determining bounds or behavior. The oscillation of \( \sin^2 k \) disrupts consistent convergence tests because it doesn't settle at a specific limit, impacting the series' convergence behavior significantly.

In calculus, trigonometric functions are often used in various contexts, from simplifying expressions to modifying series, making their comprehension essential for series analysis.