Problem 83 - Essential Calculus

Question

Determine the convergence or divergence of the series using any appropriate test from this chapter. Identify the test used. $$ \sum_{n=1}^{\infty} \frac{\cos n}{2^{n}} $$

Step-by-Step Solution

Answer

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

Step 1: Set Up the Ratio-Test

The ratio test is defined as \(\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}|\) where \({a_n}\) are the terms of our series. If the limit is less than 1, the series converges; if it is greater than 1, it diverges. If the limit equals 1 or is inconclusive, the test fails to determine convergence. Here, \(a_n = \frac{\cos n}{2^n}\). Accordingly, we need to calculate \(\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} \left|\frac{\frac{\cos (n+1)}{2^{n+1}}}{\frac{\cos n}{2^n}}\right|\).

Step 2: Simplify and Compute the Limit

We simplify \(\lim_{n \to \infty} \left|\frac{\frac{\cos (n+1)}{2^{n+1}}}{\frac{\cos n}{2^n}}\right|\) to \(\lim_{n \to \infty} \left|\frac{\cos (n+1)}{2 \cdot \cos n}\right|\). Applying the limit, as \(n\) approaches \(\infty\), \(\cos (n+1)\) and \(\cos n\) will oscillate between -1 and 1, giving no particular trend, all the while, \(2^n\) continues to increase without bound. This means the absolute fraction's value will decrease towards 0.

Step 3: Conclude the Result

Since the limit as evaluated in Step 2 is less than 1 (in this case 0), by the Ratio Test, the series \(\sum_{n=1}^{\infty} \frac{\cos n}{2^{n}}\) is convergent.

Key Concepts

Ratio Test

The ratio test is a powerful tool to determine the convergence or divergence of an infinite series. It involves evaluating the limit of the absolute value of the ratio of consecutive terms in the series. Formally, the test considers \( \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| \), where \( a_n \) represents the terms of the series.

Let's break down the steps:

• If the limit is less than 1, the series converges absolutely.
• If the limit is greater than 1 or is infinite, the series diverges.
• If the limit equals 1, the test is inconclusive.

In our example with the series \( \sum_{n=1}^{\infty} \frac{\cos n}{2^{n}} \), we used the ratio test to evaluate convergence. We found the limit to be less than 1, specifically zero, indicating the series converges. The function \( \cos n \) oscillates but is bounded, and this combined with the exponential growth of \( 2^n \) ensures the terms of the series go to zero, reinforcing the result from the ratio test.

Infinite Series

An infinite series is the sum of the terms of an infinite sequence. These series can converge to a specific value or diverge, meaning the sum does not settle on a particular number.

In mathematical terms, consider a series \( \sum_{n=1}^{\infty} a_n \), where the goal is to determine whether the sum results in a finite value (convergence) or not (divergence). Series can be classified as:

• Convergent: If the partial sums approach a finite limit as n approaches infinity.
• Divergent: If the partial sums do not approach a finite limit.

For the series given in the problem \( \sum_{n=1}^{\infty} \frac{\cos n}{2^n} \), we use convergence tests (like the ratio test) to discern its behavior. Understanding whether a series converges is crucial in analysis, enabling us to approximate or work precisely with functions represented by infinite series.

Trigonometric Series

Trigonometric series involve sine, cosine, or their combinations, and are a particular type of infinite series useful in harmonic analysis and Fourier series expansions. These series often exhibit properties based on periodic behavior and symmetry.

The series we evaluated, \( \sum_{n=1}^{\infty} \frac{\cos n}{2^{n}} \), involves cosine—a trigonometric function known for its oscillation between -1 and 1. Such functions form the basis of Fourier series, which decompose periodic functions into sums of sines and cosines.

In convergence tests like the ratio test, this oscillatory nature is significant because it directly affects the term size but is bounded. Despite the oscillation, the decreasing factor \( 2^n \) guides the convergence, as its rapid exponential growth overpowers the oscillatory factor, leading the entire term \( \frac{\cos n}{2^n} \) towards zero. Thus, understanding the nature of trigonometric functions in series helps to predict and explain convergence behaviors in mathematical analysis.