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Chapter 9: Problem 91

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

Short Answer

Expert verified
\(\Sigma(a_n / b_n)\) diverges even when \(\Sigma a_n\) and \(\Sigma b_n\) converge.

Step by step solution

01

Define the Series

We need two series, \(\Sigma a_n\) and \(\Sigma b_n\), that both converge. Let's choose \(a_n = 1/n^2\) and \(b_n = 1/(2^n)\). Both are known to converge. Specifically, \(\Sigma a_n\) is a p-series with \(p > 1\), and \(\Sigma b_n\) is a geometric series with a common ratio \(r = 1/2 < 1\).
02

Consider the Third Series

Form a new series by dividing terms: \(\frac{a_n}{b_n} = \frac{1/n^2}{1/(2^n)} = \frac{2^n}{n^2}\).
03

Check the Divergence of the Third Series

We use the limit comparison test with a known divergent series. Consider the series \(\Sigma 2^n\). The term \(\frac{2^n}{n^2}\) behaves similarly to \(2^n\) as \(n\) becomes large. This suggests divergence since \(\Sigma 2^n\) is a divergent geometric series.
04

Conclude the Divergence

Because \(\Sigma 2^n\) diverges and the terms \(\frac{2^n}{n^2}\) mimic this growth, the series \(\Sigma \frac{a_n}{b_n}\) also diverges. Each term of \(\Sigma \frac{a_n}{b_n}\) grows rapidly enough to not satisfy a finite sum.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limit Comparison Test
The Limit Comparison Test is a method used to determine the convergence or divergence of a series by comparing it to another series with known behavior. It can be particularly useful when direct convergence tests are difficult to apply. If you have two series \( \Sigma a_n \) and \( \Sigma b_n \), the test involves taking the limit: \[ L = \lim_{{n \to \infty}} \frac{a_n}{b_n}. \]
  • If \( L > 0 \) and finite, both series either converge or diverge together.
  • If \( L = 0 \) and series \( \Sigma b_n \) converges, so does \( \Sigma a_n \).
  • If \( L = \infty \) and series \( \Sigma b_n \) diverges, so does \( \Sigma a_n \).
In our exercise, it was used to compare \( \Sigma \frac{2^n}{n^2} \) with the divergent series \( \Sigma 2^n \), leading to the conclusion that \( \Sigma \frac{a_n}{b_n} \) diverges.
P-Series
A p-series is a type of infinite series defined by the reciprocal of each term raised to a power of \( p \), written as \( \Sigma \frac{1}{n^p} \). Its convergence is determined by the value of \( p \):
  • If \( p > 1 \), the series converges.
  • If \( p \leq 1 \), the series diverges.
In our example, \( \Sigma a_n \) with \( a_n = \frac{1}{n^2} \) is a p-series with \( p = 2 \), implying convergence. This choice showcases a classic example of a convergent p-series, thus establishing the known behavior against which other series can be tested.
Geometric Series
Geometric series have a distinct pattern where each term after the first is found by multiplying the previous term by a constant, called the ratio \( r \). The general form is \( \Sigma ar^n \), where \( a \) is the first term. The series converges if the absolute value of the common ratio is less than 1:
  • If \( |r| < 1 \), the series converges.
  • If \( |r| \geq 1 \), the series diverges.
In our example, \( \Sigma b_n \) with \( b_n = \frac{1}{2^n} \) is a geometric series with \( r = \frac{1}{2} \), which is less than 1, indicating that this series converges. This convergence is contrasted with the divergence of a geometric series like \( \Sigma 2^n \), where \( r > 1 \).
Convergent Series
A convergent series is a series whose sum approaches a finite number as more terms are added. To determine convergence, various tests are available, such as:
  • Comparison Test, where one checks against a known series;
  • Ratio Test, which involves the limit of term ratios;
  • Integral Test, leveraging calculus to assess series behavior.
In the given problem, both \( \Sigma a_n = \Sigma \frac{1}{n^2} \) and \( \Sigma b_n = \Sigma \frac{1}{2^n} \) demonstrate convergence, allowing us to explore how a rearrangement or combination, like \( \Sigma \frac{a_n}{b_n} \), can behave differently.
Divergent Series
A divergent series is one where the sum does not approach a finite number as more terms are added. \( \Sigma d_n \) is said to diverge if the partial sums tend to infinity or fail to settle to a particular value. Examples include:
  • Harmonic Series: \( \Sigma \frac{1}{n} \), which diverges.
  • The Series \( \Sigma 2^n \), a geometric series with a ratio greater than 1.
In our exercise, the formed series \( \Sigma \frac{a_n}{b_n} = \Sigma \frac{2^n}{n^2} \) diverges because its terms grow similarly to those in the divergent series \( \Sigma 2^n \). This confirms divergence via the Limit Comparison Test, notably when comparing with an already known divergent series.

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Most popular questions from this chapter

Which of the series, and which diverge? Give reasons for your answers. $$\sum_{n=1}^{\infty} \frac{n^{n}}{2^{\left(n^{2}\right)}}$$

Use any method to determine whether the series converges or diverges. Give reasons for your answer. $$\sum_{n=2}^{\infty} \frac{1}{1+2+2^{2}+\cdots+2^{n}}$$

Neither the Ratio Test nor the Root Test helps with \(p\) -series. Try them on $$ \sum_{n=1}^{\infty} \frac{1}{n^{p}} $$ and show that both tests fail to provide information about convergence.

Which of the series, and which diverge? Give reasons for your answers. $$\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot \cdots \cdot(2 n-1)}{[2 \cdot 4 \cdot \cdots \cdot(2 n)]\left(3^{n}+1\right)}$$

When \(a\) and \(b\) are real, we define \(e^{(a+i b) x}\) with the equation $$e^{(a+i b) x}=e^{a x} \cdot e^{i b x}=e^{a x}(\cos b x+i \sin b x)$$ Differentiate the right-hand side of this equation to show that $$\frac{d}{d x} e^{(a+i b) x}=(a+i b) e^{(a+i b) x}$$ Thus the familiar rule \((d / d x) e^{k x}=k e^{k x}\) holds for \(k\) complex as well as real.
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