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Show by example that \(\sum\left(a_{n} / b_{n}\right)\) may converge to something other than \(A / B\) even when \(A=\Sigma a_{n}, B=\Sigma b_{n} \neq 0,\) and no \(b_{n}\) equals 0.

Short Answer

Expert verified
The series \(\sum \frac{a_n}{b_n} = \sum \frac{1}{n}\) diverges despite \(\sum a_n\) converging and \(\sum b_n\) diverging.

Step by step solution

01

Define series components

Consider two series: the numerator series \(a_n\) and the denominator series \(b_n\). For this example, let's choose \(a_n = \frac{1}{n^2}\) and \(b_n = \frac{1}{n}\).
02

Calculate the sums of series

Determine the sums \(A = \sum a_n\) and \(B = \sum b_n\). The series \(a_n\) is \(\sum \frac{1}{n^2}\), which converges to \(\frac{\pi^2}{6}\). The series \(b_n\) is \(\sum \frac{1}{n}\), which diverges.
03

Formulate the new series

Form the series \(\sum \frac{a_n}{b_n} = \sum \frac{1/n^2}{1/n} = \sum \frac{1}{n}\).
04

Analyze the convergence of the new series

The new series, \(\sum \frac{1}{n}\), is the harmonic series, which is known to diverge.
05

Conclusion of example

Even though \(A = \frac{\pi^2}{6}\) and \(B\) diverges, resulting in an undefined \(\frac{A}{B}\), our constructed series \(\sum \frac{1}{n}\) diverges, showing that it does not converge to the form \(\frac{A}{B}\).

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Key Concepts

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

Harmonic Series
The harmonic series is one of the most fundamental divergent series that appears often in mathematics. This series is expressed as \( \sum \frac{1}{n} \), where \( n \) is a positive integer starting from 1 and incrementing by 1 for each following term. Each term is the reciprocal of a natural number. While the terms become very small as \( n \) increases, the harmonic series grows extraordinarily slowly yet without bound. This characteristic leads to its divergence. In our example, the new series formed was \( \sum \frac{1}{n} \), exactly the harmonic series. Despite the sum \( \sum \frac{1}{n} \) appearing to approach a finite limit due to the diminishing size of each term, it continues to grow larger indefinitely as more terms are added. This subtle property is a classic example of a series that diverges.
Convergent Series
A convergent series is one where the sum of all its terms approaches a specific finite value as more terms are added. The series \( a_n = \sum \frac{1}{n^2} \) exemplifies a convergent series, demonstrating a situation where the infinite sum of diminishing terms results in a finite number.In mathematics, for a series \( \sum a_n \) to be convergent, the limit of the partial sums of the series must exist and be finite. For example, the series \( \sum \frac{1}{n^2} \) converges to \( \frac{\pi^2}{6} \). This means as you add more terms of \( \frac{1}{n^2} \), the total sum gets closer and closer to \( \frac{\pi^2}{6} \), never exceeding this value and never failing to approach it. The convergence of a series can be determined through several tests such as the comparison test, ratio test, and integral test, which help in identifying the behavior of the series and whether it will converge or diverge.
Divergent Series
On the flip side, a divergent series is one that does not converge. In other words, the series does not approach a finite number as more terms are added. Instead, it either grows indefinitely or does not settle at any specific value.The series \( \sum \frac{1}{n} \) known as the harmonic series is a well-known divergent series. Even though its terms \( \frac{1}{n} \) keep getting smaller and smaller, the sum continues to increase forever. When talking about divergences, it is crucial to realize that a divergent series has no limit. This means attempts to assign it a fixed value would not reflect the true behavior of the series. In mathematical terms, no matter how many terms you add up in the series, it will never approach a specific number, showing that it is divergent and its sum is ultimately infinite.

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Most popular questions from this chapter

The sum of the series \(\sum_{n=0}^{\infty}\left(n^{2} / 2^{n}\right)\) To find the sum of this series, express \(1 /(1-x)\) as a geometric series, differentiate both sides of the resulting equation with respect to \(x,\) multiply both sides of the result by \(x\), differentiate again, multiply by \(x\) again, and set \(x\) equal to \(1 / 2 .\) What do you get?

Use any method to determine whether the series converges or diverges. Give reasons for your answer. $$\sum_{n=0}^{\infty}(-1)^{n} \frac{(n+2) !}{(2 n) !}$$

In the alternating harmonic series, suppose the goal is to arrange the terms to get a new series that converges to \(-1 / 2 .\) Start the new arrangement with the first negative term, which is \(-1 / 2 .\) Whenever you have a sum that is less than or equal to \(-1 / 2,\) start introducing positive terms, taken in order, until the new total is greater than \(-1 / 2 .\) Then add negative terms until the total is less than or equal to \(-1 / 2\) again. Continue this process until your partial sums have been above the target at least three times and finish at or below it. If \(s_{n}\) is the sum of the first \(n\) terms of your new series, plot the points \(\left(n, s_{n}\right)\) to illustrate how the sums are behaving.

Prove that if \(\sum a_{n}\) converges absolutely, then \(\sum a_{n}^{2}\) converges.

Use any method to determine whether the series converges or diverges. Give reasons for your answer. $$\sum_{n=1}^{\infty} \frac{4 \cdot 6 \cdot 8 \cdots(2 n)}{5^{n+1}(n+2) !}$$

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