Problem 14 - Calculus: Early Transcendentals

Question

Test the series for convergence or divergence. \( \displaystyle \sum_{n = 1}^{\infty} \frac {\sin 2n}{1 + 2^n} \)

Step-by-Step Solution

Answer

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

Step 1: Identify the Series Type

The series given is \( \sum_{n=1}^{\infty} \frac{\sin 2n}{1 + 2^n} \). It is an infinite series where each term is \( \frac{\sin 2n}{1 + 2^n} \). We are asked to determine whether the series converges or diverges.

Step 2: Analyze the General Term

The general term is \( a_n = \frac{\sin 2n}{1 + 2^n} \). Note that \( \sin 2n \) is a bounded function that oscillates between -1 and 1. Therefore, \( |\sin 2n| \leq 1 \). Since the denominator \( 1 + 2^n \) grows exponentially, the entire fraction approaches zero as \( n \to \infty \).

Step 3: Apply the Limit Comparison Test

Find a simple series to compare with. Consider \( b_n = \frac{1}{2^n} \), which is a geometric series with a common ratio of \( \frac{1}{2} \). This geometric series converges since its common ratio is less than 1.

Step 4: Compare Both Series

We compare \( a_n \) with \( b_n \): \[\lim_{{n \to \infty}} \frac{a_n}{b_n} = \lim_{{n \to \infty}} \frac{\frac{\sin 2n}{1 + 2^n}}{\frac{1}{2^n}} = \lim_{{n \to \infty}} \frac{\sin 2n \cdot 2^n}{1 + 2^n}. \]Rewriting the limit expression, this becomes:\[ \lim_{{n \to \infty}} \frac{\sin 2n \cdot 2^n}{1 + 2^n} = \lim_{{n \to \infty}} \frac{\sin 2n}{\frac{1}{2^n} + 1}. \] Since \( |\sin 2n| \leq 1 \), the term in the denominator \( \frac{1}{2^n} + 1 \) approaches 1 as \( n \to \infty \). Hence,\[\lim_{{n \to \infty}} \frac{\sin 2n}{\frac{1}{2^n} + 1} = \frac{1}{1} = 1.\]Since the limit is finite and nonzero, by the Limit Comparison Test, since \( \sum b_n \) converges, \( \sum a_n \) also converges.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a powerful tool used in analyzing the convergence of infinite series. It is particularly useful when comparing complex series to simpler, well-understood series. The main idea is to compare the general terms of two series to determine if they behave similarly as they approach infinity.

### How It Works

• Identify two series: one is the series of interest, say \(a_n\), and the other is a simpler series, say \(b_n\), for which you already know whether it converges or diverges.
• Calculate the limit \( L = \lim_{{n o \infty}} \frac{a_n}{b_n} \).
• Conclude based upon this limit:
  
  • If \(L\) is positive and finite (i.e., 0 < \(L\) < ∞), then both series either converge or diverge together.
  • If \(L = 0\) and \(b_n\) converges, then \(a_n\) also converges.
  • If \(L = \infty\) and \(b_n\) diverges, then \(a_n\) also diverges.

The Limit Comparison Test helps simplify complex series by relating them to a more straightforward series. In our given problem, it helped demonstrate the convergence of a complex trigonometric series by comparing it to a geometric series.

Infinite Series

An infinite series is essentially the sum of the terms of an infinite sequence. It is expressed as \( \sum_{n=1}^{\ackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslash\ackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslashackslash\infty} a_n\), where \(a_n\) represents the general terms of the series. Understanding whether an infinite series converges or diverges is crucial, as it helps determine if the series sums to a specific number or grows indefinitely.

### Convergence vs. Divergence

• **Convergent Series:** An infinite series converges if the series has a finite limit as the number of terms \approaches ∞.
• **Divergent Series:** An infinite series diverges if the limit doesn't exist or is infinite.

To determine convergence, various tests like the Limit Comparison Test, Ratio Test, and others can be employed. Our problem focused on using the Limit Comparison Test to demonstrate convergence.

Geometric Series

A geometric series is a type of infinite series where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the "common ratio." Geometric series are simpler and easier to analyze compared to other complex series.

### Mathematical Formula

• A geometric series is expressed as \( \sum_{n=0}^{\ackslashackslashackslashackslashackslashackslashackslash\infty} ar^n\), where \(a\) is the initial term and \(r\) is the common ratio.

### Convergence Condition

• The series converges if the common ratio \(|r| < 1\).
• The series is divergent if \(|r| \geq 1\).

In the given exercise, the series we compared with was a geometric series \(b_n = \frac{1}{2^n}\), which converges because its common ratio \(r = \frac{1}{2}\) is less than one.

General Term Analysis

General term analysis is a crucial step in understanding the behavior of series and sequences. It involves examining the form of individual terms in a series \(a_n\) to make deductions regarding convergence or divergence.

### Key Observations

• For oscillating functions like \(\sin(2n)\), the values are bounded between -1 and 1, providing insight into the term's bounds.
• The exponential terms like \(1 + 2^n\) in the denominator grow very fast, influencing the rate at which \(a_n\) approaches zero.

By analyzing these generalized terms, we deduce how the series behaves as \(n\) approaches infinity. In the given exercise, understanding these dynamics allowed us to determine that \(\frac{\sin 2n}{1 + 2^n}\) approaches zero, aiding in the application of the Limit Comparison Test for convergence.