Problem 11 - Thomas Calculus

Question

In Exercises \(9-16,\) use the Root Test to determine if each series converges absolutely or diverges. $$\sum_{n=1}^{\infty}\left(\frac{4 n+3}{3 n-5}\right)^{n}$$

Step-by-Step Solution

Answer

The series diverges by the Root Test as \(L = \frac{4}{3} > 1\).

Step 1: Understand the Root Test

The Root Test helps determine the convergence of a series \(\sum_{n=1}^{\infty} a_n\). According to the Root Test, calculate \(L = \lim_{n \to \infty} \sqrt[n]{|a_n|}\). The series converges absolutely if \(L < 1\); diverges if \(L > 1\); and if \(L = 1\), the test is inconclusive.

Step 2: Determine \(a_n\)

In this problem, the series given is \(\sum_{n=1}^{\infty} \left( \frac{4n+3}{3n-5} \right)^n\). Thus, we have \(a_n = \left( \frac{4n+3}{3n-5} \right)^n\).

Step 3: Apply the Root Test

To use the Root Test, find the \(n\)-th root of \(|a_n|\), which is \(|a_n|^{1/n}\). We have \(\left( \left( \frac{4n+3}{3n-5} \right)^n \right)^{1/n}\).

Step 4: Simplify the Expression

Simplify the expression \( \left( \left( \frac{4n+3}{3n-5} \right)^n \right)^{1/n} = \frac{4n+3}{3n-5} \).

Step 5: Take the Limit

Calculate \(L = \lim_{n \to \infty} \frac{4n+3}{3n-5}\). As \(n\) becomes large, the leading terms dominate, so \( \lim_{n \to \infty} \frac{4n+3}{3n-5} = \lim_{n \to \infty} \frac{4n}{3n} = \frac{4}{3}\).

Step 6: Conclude Using the Root Test

Since \(L = \frac{4}{3} > 1\), by the Root Test, the series \(\sum_{n=1}^{\infty} \left( \frac{4n+3}{3n-5} \right)^n\) diverges.

Key Concepts

series convergence

When we talk about series convergence, we are discussing whether the sum of an infinite series, such as \( \sum_{n=1}^{\infty} a_n \), adds up to a finite number or not. If the series adds up to a definite, fixed number, we say it converges. Otherwise, it diverges. Understanding series convergence is crucial because it helps us predict the behavior of series in mathematics.
For example, when we use tests like the Root Test, we assess if a series converges absolutely, which means even the series of absolute values converges. Identifying convergence ensures the series is well-behaved in mathematical applications.

absolute convergence

The idea of absolute convergence is an extension of series convergence. A series \( \sum_{n=1}^{\infty} a_n \) is said to converge absolutely if the series of absolute values \( \sum_{n=1}^{\infty} |a_n| \) converges. This implies that not only does the series itself converge, but it would still converge even if all terms were made positive.
An absolutely convergent series is always convergent. However, the reverse—convengent series being absolutely convergent—is not necessarily true. Absolute convergence assures us of a stronger form of convergence that guarantees stability in the sum across all scenarios.

limit comparison

Limit comparison is a technique often used when testing convergence of series. When direct application of certain tests is impractical, we compare the series with another known series. Here, we use the limit comparison test: Suppose we have two series \( \sum_{n=1}^{\infty} a_n \) and \( \sum_{n=1}^{\infty} b_n \) such that \( a_n > 0 \) and \( b_n > 0 \) for all \( n \).
By taking \( L = \lim_{n \to \infty} \frac{a_n}{b_n} \), we can determine the convergence of \( \sum a_n \) based on that of \( \sum b_n \):

• If \( 0 < L < \infty \), both series converge or both diverge.
• If \( L = 0 \) and \( \sum b_n \) converges, so does \( \sum a_n \).
• If \( L = \infty \) and \( \sum b_n \) diverges, so does \( \sum a_n \).

This test is powerful because it allows us to infer the behavior of complex series with simpler comparisons.