Problem 14 - Calculus Single Variable

Question

Use the Comparison Test for Convergence to show that the given series converges. State the series that you use for comparison and the reason for its convergence. $$ \sum_{n=1}^{\infty} \frac{2}{2^{n}+1 / 2^{n}} $$

Step-by-Step Solution

Answer

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

Step 1: Simplify the Series

First, simplify the series to make it easier to compare. The given series is \( \sum_{n=1}^{\infty} \frac{2}{2^{n} + 1 / 2^{n}} \). Recognize that \( 1 / 2^{n} \) is much smaller than \( 2^{n} \) as \( n \) grows, so the denominator can be approximated as \( 2^{n} \). Thus, \( \frac{2}{2^{n} + 1 / 2^{n}} \approx \frac{2}{2^{n}} = \frac{1}{2^{n-1}} \).

Step 2: Identify the Comparison Series

Choose a simpler series to compare with. Based on the simplification, we use the geometric series \( \sum_{n=1}^{\infty} \frac{1}{2^n} \).

Step 3: Check for Convergence of the Comparison Series

The series \( \sum_{n=1}^{\infty} \frac{1}{2^n} \) is a geometric series with a common ratio \( r = \frac{1}{2} \) (less than 1), which is known to converge.

Step 4: Applying the Comparison Test

Since \( \frac{2}{2^{n} + 1 / 2^{n}} \leq \frac{1}{2^{n-1}} \) for all \( n \), and \( \sum_{n=1}^{\infty} \frac{1}{2^{n-1}} = \sum_{n=1}^{\infty} \frac{1}{2} \left( \frac{1}{2^n} \right) \) which is a convergent series, by the Comparison Test the original series \( \sum_{n=1}^{\infty} \frac{2}{2^{n}+1 / 2^{n}} \) also converges.

Key Concepts

Convergence of Series

Convergence of a series is a fundamental concept in calculus. When we say a series converges, it means that as we sum its infinite number of terms, the total approaches a finite number. In mathematical terms, if the series \( \sum_{n=1}^{\infty} a_n \) converges, the sequence of partial sums \( S_n = a_1 + a_2 + \ldots + a_n \) approaches a limit as \( n \to \infty \). This concept helps us analyze the behavior of series and determine if they settle at a specific value or go to infinity.

There are various tests to check for convergence, such as the Comparison Test, Ratio Test, and more. These tests provide different approaches to determine whether we reach a finite number by adding up series' terms. For the problem given, we employed the Comparison Test.

• We simplified the original series to make it more understandable.
• We then chose another simpler series to compare which already is known to converge.

By confirming the simpler series converges and checking the conditions of the Comparison Test, the original series was shown to converge as well. This confirms its behavior and tells us important information about how it behaves when summed to infinity.

Geometric Series

A geometric series is a type of series where each term is a constant multiple of the previous term. The basic form can be expressed as \( \sum_{n=0}^{\infty} ar^n \), where \( a \) is the first term and \( r \) is the common ratio.

Geometric series are particularly important because they are one of the few series types with straightforward convergence conditions. A geometric series converges if the absolute value of the common ratio \( |r| \) is less than 1.

• If \( |r| < 1 \), the series converges to \( \frac{a}{1-r} \).
• If \( |r| \geq 1 \), the series diverges.

In the original problem, the comparison involved a geometric series \( \sum_{n=1}^{\infty} \frac{1}{2^n} \), which has a common ratio \( r = \frac{1}{2} \). Because \( \frac{1}{2} < 1 \), this series converges. This gives a strong foundational example of how to apply concepts of geometric series and helps justify using them for comparison tests in other convergence problems.

Series Approximation

Series approximation is a handy technique that enables us to make complex series simpler and easier to manage. By simplifying or approximating the expressions within a series, we can often identify known series forms to compare against, such as the geometric series mentioned earlier.

In the exercise provided, approximating the original series helped immensely. Initially, the series expression seemed complicated, but by understanding that for large \( n \), \( \frac{1}{2^n} \) becomes negligible and the expression \( 2^n \) dominates, the series was approximated to look like \( \frac{1}{2^{n-1}} \). This is simpler to handle and resembles the form of a geometric series.

• Approximation helped identify a comparable known series.
• We could then use convergence knowledge of simpler series for reasoning.

Through series approximation, we streamlined the process of determining convergence by relating complex series to more familiar and manageable forms, making the analysis much easier.