Chapter 11: Problem 63
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 \(0 .\)
Short Answer
Expert verified
The series \(\sum \frac{3^n}{2^n}\) diverges, illustrating that \(\sum \frac{a_n}{b_n}\) can diverge even if \(\sum a_n\) and \(\sum b_n\) converge.
Step by step solution
01
Understand the Problem
We are given two series \(\sum a_n\) and \(\sum b_n\) which are both convergent. We need to show that the series \(\sum \frac{a_n}{b_n}\) can diverge even if both individual series \(\sum a_n\) and \(\sum b_n\) converge. Additionally, it is given that \(b_n eq 0\) for all \(n\).
02
Example Series Setup
To construct our example, let's choose \(a_n = \frac{1}{2^n}\) and \(b_n = \frac{1}{3^n}\). Both \(\sum a_n = \sum \frac{1}{2^n}\) and \(\sum b_n = \sum \frac{1}{3^n}\) are geometric series with ratios less than 1, and hence are convergent.
03
Construct the Series \(\sum \frac{a_n}{b_n}\)
Now, consider the series \(\sum \frac{a_n}{b_n} = \sum \frac{\frac{1}{2^n}}{\frac{1}{3^n}} = \sum \left( \frac{3^n}{2^n} \right)\). This simplifies to \(\left(\frac{3}{2}\right)^n\).
04
Test for Divergence
The series \(\sum \left(\frac{3}{2}\right)^n\) is a geometric series with ratio \(\frac{3}{2} > 1\). A geometric series diverges if its common ratio is greater than 1. Therefore, \(\sum \left(\frac{3}{2}\right)^n\) diverges.
05
Conclusion
We have shown through the example that \(\sum \frac{a_n}{b_n}\) can indeed diverge even though \(\sum a_n\) and \(\sum b_n\) both converge. Hence, the given statement is true.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Convergent Series
A convergent series is one where the sum of its terms approaches a specific finite number, known as the limit, as more and more terms are added. Convergence happens when the partial sums of the series become closer and closer to a certain value. This means that as you take the sum of a very large number of terms, the result gets very close to a certain number rather than infinitely increasing or decreasing. For example, the series \(\sum \frac{1}{2^n}\) is convergent because each additional term \(\frac{1}{2^n}\) gets smaller and smaller, causing the overall sum to approach a fixed value.
Some key points about convergent series:
Some key points about convergent series:
- They have a finite sum.
- Adding more terms brings the sum closer to a certain number.
- Geometric series with a ratio \(r < 1\) are examples of convergent series.
Geometric Series
A geometric series is a series with a constant ratio between consecutive terms. The general form of a geometric series is \(a + ar + ar^2 + ar^3 + \ldots\), where \(a\) is the first term and \(r\) is the common ratio.
Geometric series can either converge or diverge, depending on the value of the common ratio \(r\):
Geometric series can either converge or diverge, depending on the value of the common ratio \(r\):
- If \(|r| < 1\), the series converges and the sum can be calculated using the formula \(S = \frac{a}{1-r}\).
- If \(|r| \geq 1\), the series diverges, meaning the sum is not finite.
Series Divergence
A series is divergent if its terms do not approach any finite sum as more and more terms are added. Instead of getting closer to a specific value, the sum grows indefinitely or oscillates without settling into a finite limit. This happens when the partial sums of a series keep increasing or never stabilize to one value when added infinitely.
Key signs of divergence include:
Key signs of divergence include:
- The partial sums increase without bound.
- The terms of the series do not get progressively smaller.
- For geometric series, if the ratio \(r\) is \(|r| \geq 1\), the series diverges.
Ratio Test
The ratio test is a handy tool to determine whether a series is convergent or divergent. This test involves calculating the limit of the absolute value of the ratio of a term and the next term in the series. Suppose \(a_n\) is a term in the series; we consider the series \(\sum a_n\):
- Compute \(L = \lim_{n \to \infty}\left| \frac{a_{n+1}}{a_n} \right|\).
- If \(L < 1\), the series \(\sum a_n\) converges.
- If \(L > 1\), the series \(\sum a_n\) diverges.
- If \(L = 1\), the test is inconclusive, meaning other methods are required.