Problem 83 - Thomas Calculus

Question

Show by example that \(\sum\left(a_{n} / b_{n}\right)\) may diverge even though \(\Sigma a_{n}\) and \(\sum b_{n}\) converge and no \(b_{n}\) equals \(0 .\)

Step-by-Step Solution

Answer

Example: \( a_n = \frac{1}{2^n}, b_n = \frac{1}{3^n} \). \( \sum \frac{a_n}{b_n} = \sum \left( \frac{3}{2} \right)^n \) diverges.

Step 1: Understand the Problem

We are given two series \( \sum a_n \) and \( \sum b_n \) that both converge, and we need to show that their ratio series \( \sum \left( \frac{a_n}{b_n} \right) \) can diverge. Importantly, each term \(b_n eq 0 \).

Step 2: Choose Example Series

Consider \(a_n = \frac{1}{2^n}\) and \(b_n = \frac{1}{3^n}\). Both of these series are geometric series, and we know geometric series converge when their common ratio is less than 1.

Step 3: Check Convergence of \(\sum a_n \)

The series \( \sum a_n = \sum \frac{1}{2^n} \) is a geometric series with common ratio \( \frac{1}{2} \), which is less than 1, hence, it converges.

Step 4: Check Convergence of \(\sum b_n \)

The series \( \sum b_n = \sum \frac{1}{3^n} \) is a geometric series with common ratio \( \frac{1}{3} \), which is less than 1, hence, it converges.

Step 5: Construct \(\sum \left( \frac{a_n}{b_n} \right) \) and Simplify

We have \( \frac{a_n}{b_n} = \frac{\frac{1}{2^n}}{\frac{1}{3^n}} = \left(\frac{3}{2}\right)^n \).

Step 6: Check Convergence of \(\sum \left( \frac{a_n}{b_n} \right) \)

Now, consider the series \( \sum \left( \frac{3}{2} \right)^n \). This series is a geometric series with a common ratio of \( \frac{3}{2} \), which is greater than 1. Therefore, this series diverges.

Step 7: Conclusion and Verification

Since both \( \sum a_n \) and \( \sum b_n \) converge but \( \sum \left( \frac{a_n}{b_n} \right) \) diverges, we have shown by this example that the statement is true.

Key Concepts

Geometric Series

A geometric series is a type of series where each term is a fixed multiple of the previous one. This fixed multiple is known as the 'common ratio'. For a series to be geometric, the sequence of terms should form a progression by multiplication.
For example, a series like \( a, ar, ar^2, ar^3, \cdots \) is geometric, where \( a \) is the first term and \( r \) is the common ratio.

• If the absolute value of the common ratio \( |r| \) is less than 1, the series converges, meaning that as more terms are added, the sum approaches a finite number.
• If \( |r| \) is equal to or greater than 1, the series diverges, and the sum will grow indefinitely as more terms are added.

In the exercise, both series \( \sum a_n \) and \( \sum b_n \) are geometric with \( r = \frac{1}{2} \) and \( r = \frac{1}{3} \) respectively, so they both converge.

Convergence

Convergence is a fundamental concept in calculus and the study of series. It refers to the behavior of a series as the number of terms increases.
When we say a series converges, it means the sum of its terms approaches a specific value, even if an infinite number of terms are added.
For a series \( \sum a_n \), it converges if \( \lim_{n \to \infty} S_n = L \), where \( S_n \) is the partial sum of the first n terms, and \( L \) is some finite limit. In contrast, if the series does not converge, it is said to diverge, meaning the partial sums do not approach any finite value as more terms are added.
Once convergence is established, one might naturally assume that certain operations, such as taking the ratio of two converging series, would also lead to convergence. However, this exercise shows that even when two series individually converge, their term-wise ratio \( \sum \left( \frac{a_n}{b_n} \right) \) can still diverge as demonstrated by expanding each term.

Ratio Test

The ratio test is a common method for determining the convergence or divergence of an infinite series.
To apply the ratio test, consider a series \( \sum a_n \) and evaluate the limit of the absolute value of the ratio of consecutive terms:
\[L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]

• If \( L < 1 \), the series converges.
• If \( L > 1 \) or \( L \) is infinite, the series diverges.
• If \( L = 1 \), the test is inconclusive, and the series may converge or diverge; further analysis would be needed.

In the example from the exercise, applying the ratio test to \( \sum \left( \frac{3}{2} \right)^n \) confirms divergence:
The common ratio \( \frac{3}{2} \) results in \( L = \frac{3}{2} > 1 \), indicating divergence. This demonstrates how powerful the ratio test can be in analyzing the behavior of series.