Problem 19 - Thomas Calculus

Question

Which of the series in Exercises \(17-56\) converge, and which diverge? Use any method, and give reasons for your answers. $$ \sum_{n=1}^{\infty} \frac{\sin ^{2} n}{2^{n}} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Series

The series we need to evaluate is \( \sum_{n=1}^{\infty} \frac{\sin^2 n}{2^n} \). We need to determine if this series converges or diverges.

Step 2: Determine the Type of Series

Observe that the series is of the form \( \sum_{n=1}^{\infty} a_n \) where \( a_n = \frac{\sin^2 n}{2^n} \). Here, \( \sin^2 n \) is always positive and bounded between 0 and 1, and the denominator is an exponential function of base 2.

Step 3: Apply the Comparison Test

Since \( 0 \leq \sin^2 n \leq 1 \) for all \( n \), each term \( a_n = \frac{\sin^2 n}{2^n} \) satisfies \( 0 \leq a_n \leq \frac{1}{2^n} \). We know that the geometric series \( \sum_{n=1}^{\infty} \frac{1}{2^n} \) converges because it is a geometric series with common ratio \( \frac{1}{2} \), which is less than 1.

Step 4: Conclude Using the Comparison Test

Since \( 0 \leq \frac{\sin^2 n}{2^n} \leq \frac{1}{2^n} \) and the series \( \sum_{n=1}^{\infty} \frac{1}{2^n} \) converges, by the Comparison Test, the series \( \sum_{n=1}^{\infty} \frac{\sin^2 n}{2^n} \) also converges.

Key Concepts

Comparison Test

The Comparison Test is a useful method to determine the convergence or divergence of a given series. When you have a series and need to figure out its behavior, the Comparison Test comes in handy by comparing it with another series whose convergence properties are already known.
For the Comparison Test, we need two series: one from the exercise (let's call it \( \sum a_n \)) and a reference series (\( \sum b_n \)). The idea is pretty simple:

• If \( 0 \leq a_n \leq b_n \) for all \( n \) (from a certain point), and the reference series \( \sum b_n \) converges, then the series \( \sum a_n \) also converges.
• Conversely, if \( a_n \geq b_n \geq 0 \) for all \( n \), and \( \sum b_n \) diverges, then \( \sum a_n \) diverges too.

In our original exercise, we used the Comparison Test by noting that for our series, \( a_n = \frac{\sin^2 n}{2^n} \) and comparing it to the geometric series \( \sum \frac{1}{2^n} \). This helped us determine that the original series converges.
It's always crucial to ensure that the conditions of the test are met, especially the non-negativity of terms.

Geometric Series

A geometric series is a series of the form \( \sum_{n=0}^{\infty} ar^n \), where \( a \) is the first term, and \( r \) is the common ratio between consecutive terms.
One major property of geometric series is the rule for convergence. A geometric series converges if the absolute value of the common ratio \( r \) is less than 1. Otherwise, if \( |r| \geq 1 \), the series diverges.

• If the series \( \sum ar^n \) converges, its sum can be calculated using the formula \( S = \frac{a}{1-r} \).
• Familiarity with geometric series can be quite advantageous in analyzing other series through comparison.

For instance, the series \( \sum \frac{1}{2^n} \) in our example is geometric with \( r = \frac{1}{2} \), which is less than 1. Therefore, it converges.
This property makes geometric series a valuable reference in the Comparison Test to infer the behavior of other series.

Convergence and Divergence

Understanding convergence and divergence is key in the study of infinite series. These concepts tell us whether a series adds up to a finite value or not as more terms are included.
Convergence is when the sum of a series approaches a specific number. Divergence, on the other hand, means the sum does not settle at any finite number and tends to infinity or oscillates indefinitely.

• Convergence indicates that we can approximate the sum to a fixed, finite value.
• Divergence often implies that further scrutiny or a different approach might be needed since summing the sequence doesn't yield useful numerical information.

It is crucial to apply appropriate tests, like the Comparison Test, to assess convergence or divergence. In our series \( \sum \frac{\sin^2 n}{2^n} \), showing it is bounded by a known convergent series confirms its convergence.
Remember, using systematic tests to infer the behavior of a series helps ensure correctness in results and clarity in understanding, especially when initial inspection is daunting.