Problem 42 - Calculus Early Transcendentals: Pearson New International Edition

Question

Test for convergence or divergence using the Root Test. (a) \(\sum_{n=2}^{\infty}\left(\frac{1}{\ln n}\right)^{n}\) (b) \(\sum_{n=1}^{\infty}\left(\frac{n}{3 n+2}\right)^{n}\) (c) \(\sum_{n=1}^{\infty}\left(\frac{1}{2}+\frac{1}{n}\right)^{n}\)

Step-by-Step Solution

Answer

All series (a), (b), and (c) converge using the Root Test.

Step 1: Understanding the Root Test

The Root Test, also known as the Cauchy Root Test, is a method used to determine the convergence or divergence of an infinite series \( \sum a_n \). The test is based on evaluating the limit \( \lim_{n \to \infty} \sqrt[n]{|a_n|} = L \). If \( L < 1 \), the series converges; if \( L > 1 \), the series diverges; if \( L = 1 \), the test is inconclusive.

Step 2: Test for Convergence of Series (a)

Consider the series \( \sum_{n=2}^{\infty}\left(\frac{1}{\ln n}\right)^{n} \). Here, the general term is \( a_n = \left(\frac{1}{\ln n}\right)^{n} \). Apply the Root Test: calculate \( \lim_{n \to \infty} \sqrt[n]{\left(\frac{1}{\ln n}\right)^{n}} = \lim_{n \to \infty} \frac{1}{\ln n} \). As \( n \to \infty \), \( \ln n \to \infty \), thus \( \lim_{n \to \infty} \frac{1}{\ln n} = 0 \). Since the limit is less than 1, the series converges.

Step 3: Test for Convergence of Series (b)

Consider the series \( \sum_{n=1}^{\infty}\left(\frac{n}{3n+2}\right)^{n} \). The general term is \( a_n = \left(\frac{n}{3n+2}\right)^{n} \). Apply the Root Test: \( \lim_{n \to \infty} \sqrt[n]{\left(\frac{n}{3n+2}\right)^{n}} = \lim_{n \to \infty} \frac{n}{3n+2} \). This simplifies to \( \lim_{n \to \infty} \frac{1}{3}, \) which is \( \frac{1}{3} < 1 \). Hence, the series converges.

Step 4: Test for Convergence of Series (c)

Consider the series \( \sum_{n=1}^{\infty}\left(\frac{1}{2}+\frac{1}{n}\right)^{n} \). The general term is \( a_n = \left(\frac{1}{2}+\frac{1}{n}\right)^{n} \). Apply the Root Test: \( \lim_{n \to \infty} \sqrt[n]{\left(\frac{1}{2}+\frac{1}{n}\right)^{n}} = \lim_{n \to \infty} \left(\frac{1}{2}+\frac{1}{n}\right) \). As \( n \to \infty \), \( \left(\frac{1}{2}+\frac{1}{n}\right) \to \frac{1}{2} \). Since \( \frac{1}{2} < 1 \), the series converges.

Key Concepts

Infinite Series

An infinite series is a sum of an infinite sequence of terms. In mathematical notation, we often write it as \( \sum_{n=1}^{\infty} a_n \), where each \( a_n \) represents a term in the sequence. Unlike a finite series, where we add up a limited number of terms, infinite series continue indefinitely. Depending on the behavior of their terms, these series can either "converge" or "diverge."

• Converge: If the sum of the series approaches a particular finite value as more and more terms are added, we say the series converges.
• Diverge: Conversely, if the sum doesn’t approach a fixed number, or it increases without bound, the series diverges.

Understanding the nature of an infinite series helps us determine its behavior as we examine more terms. This forms the basis for understanding complex mathematical concepts, like convergence and divergence, which are critical in calculus and analysis.

Convergence and Divergence

Convergence and divergence describe the behavior of infinite series and sequences. These concepts help in determining the limit to which a series approaches or if it veers off towards infinity.Convergence of a SeriesWhen a series converges, the sum of its terms approaches a specific finite limit as more terms are added. For example, in the harmonic series or a geometrical progression with a common ratio \(r < 1\), the series converges toward a limit. Consider a simple example with a geometric progression:\( 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \)It converges to \( 2 \).Divergence of a SeriesOn the other hand, a divergent series doesn't settle at any finite limit. Its terms grow too large, or the sum oscillates without approaching a particular value. For instance, a series like\( 1 + 1 + 1 + 1 + \ldots \) will diverge because adding more ones continues to increase the overall sum indefinitely.Being able to identify whether a series converges or diverges is key in various mathematical problems and real-world applications.

Cauchy Root Test

The Cauchy Root Test is a useful tool for determining the convergence or divergence of an infinite series. It's named after the mathematician Augustin-Louis Cauchy. This test is beneficial because it simplifies the calculation process for many series.**Root Test Details**The test involves taking the nth root of the absolute value of the series' general term, \( a_n \). The idea is to find:\[ \lim_{n \to \infty} \sqrt[n]{|a_n|} = L \]Once you determine \( L \), the convergence is decided as follows:

• If \( L < 1 \), the series converges.
• If \( L > 1 \), the series diverges.
• If \( L = 1 \), the test is inconclusive; other methods must be used to determine the behavior of the series.

In simple terms, if the terms of the series shrink quickly enough to zero, the series will likely converge. If they don't shrink quickly, or increase, it likely diverges.The Cauchy Root Test is particularly useful for series with exponential or factorial terms, as these often yield clear results when subjected to nth root evaluation.