Problem 19 - Thomas Calculus

Question

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \frac{\sin ^{2} n}{2^{n}}\end{equation}

Step-by-Step Solution

Answer

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

Step 1: Identify the Series Type

The given series is \( \sum_{n=1}^{\infty} \frac{\sin^2 n}{2^n} \). This is an infinite series where each term is of the form \( \frac{\sin^2 n}{2^n} \).

Step 2: Recognize the Range of the Sine Function

Note that \( 0 \leq \sin^2 n \leq 1 \) for any integer \( n \). This is because sine values range from -1 to 1, and squaring these values results in a range from 0 to 1.

Step 3: Compare with a Known Convergent Series

We can compare this series to a geometric series. Since \( 0 \leq \frac{\sin^2 n}{2^n} \leq \frac{1}{2^n} \), consider the series \( \sum_{n=1}^{\infty} \frac{1}{2^n} \), which is a geometric series with common ratio \( r = \frac{1}{2} \).

Step 4: Check the Convergence of the Comparison Series

The geometric series \( \sum_{n=1}^{\infty} \frac{1}{2^n} \) converges, because the common ratio \( r = \frac{1}{2} < 1 \). Its sum is \( 1 \).

Step 5: Apply 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

Infinite Series

An infinite series is a sum of an infinite sequence of terms. If you imagine adding numbers forever, you're thinking of an infinite series! These series are given by expressions like \( \sum_{n=1}^{\infty} a_n \), where \( a_n \) stands for each term in the sequence. Infinite series can either converge or diverge. A series converges if the sum approaches a finite limit as more and more terms are added. Conversely, it diverges if the sum keeps growing indefinitely or oscillates without settling towards a single value. Many approaches help determine this behavior, each suited to different series types.

Geometric Series

A geometric series is a series where each term is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio. The general form of a geometric series is \( a + ar + ar^2 + ar^3 + \ldots \). For an infinite geometric series, this can be represented as \( \sum_{n=0}^{\infty} ar^n \). The series converges when the absolute value of the common ratio \( |r| \) is less than 1. If it converges, its sum can be calculated with the formula \( S = \frac{a}{1 - r} \), where \( a \) is the first term of the series. Geometric series are particularly useful in the study of convergence because they provide a simple way to compare other series, as seen in the original solution.

Comparison Test

The comparison test is a handy tool to determine the convergence or divergence of an infinite series. This method involves comparing the series of interest with a second series that has a known behavior (whether it converges or diverges).

• If the series of interest has terms that are smaller (or less than or equal to) terms of a known convergent series, then it also converges.
• On the other hand, if its terms are larger (or greater than or equal to) terms of a known divergent series, it must diverge.

This is because the relative size of terms can provide evidence of the overall behavior of the series. In our exercise, we used the comparison test by pairing the given series with a known geometric series, leading to the conclusion that the original series converges.