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Chapter 8: Problem 10

Give an example of a series that converges conditionally but not absolutely.

Short Answer

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Question: Give an example of a series that converges conditionally but not absolutely and explain why it converges conditionally and not absolutely. Answer: One example of a series that converges conditionally but not absolutely is the alternating harmonic series, given by the formula \(\sum\limits_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n}\). It converges conditionally because it satisfies the conditions of the Alternating Series Test, as the sequence of absolute values is non-increasing and converges to zero. However, it does not converge absolutely, as the sum of the absolute values of its terms is the harmonic series, which is known to diverge.

Step by step solution

01

Series Example

One example of a series that converges conditionally but not absolutely is the alternating harmonic series, which is given by the following formula: \( \sum\limits_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n}\).
02

Conditional Convergence

To show that the alternating harmonic series converges, we can use the Alternating Series Test. The Alternating Series Test states that if the following two conditions are satisfied, then the series converges: 1. The sequence of absolute values is non-increasing, i.e., \(|a_{n+1}| \leq |a_n|\) for all \(n\). 2. The sequence of absolute values converges to zero, i.e., \(\lim\limits_{n\to\infty} |a_n| = 0\). For the alternating harmonic series, we can easily show that both of these conditions are satisfied: 1. The sequence of absolute values is \(\frac{1}{n}\), which is non-increasing, since \(\frac{1}{n+1} \leq \frac{1}{n}\) for all \(n\). 2. The limit of the absolute values is \(\lim\limits_{n\to\infty} \frac{1}{n} = 0\). Since both conditions are satisfied, the alternating harmonic series converges conditionally.
03

Absolute Convergence

Now we need to show that the alternating harmonic series does not converge absolutely. To do this, we need to find the sum of the absolute values of the terms: \( \sum\limits_{n=1}^{\infty} |(-1)^{n+1} \frac{1}{n}| = \sum\limits_{n=1}^{\infty} \frac{1}{n}\). This is the harmonic series, which is a well-known example of a divergent series. It can be shown that the harmonic series diverges using the Integral Test or the Comparison Test. Since the harmonic series diverges, the alternating harmonic series does not converge absolutely. Our example, the alternating harmonic series, converges conditionally but not absolutely.

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Key Concepts

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

Alternating Harmonic Series
The alternating harmonic series is a captivating example in the study of infinite series. This series is defined as:\[\sum\limits_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n}\]What makes this series unique is its alternating signs, which means that consecutive terms switch between positive and negative. The pattern looks like: - 1 - \(\frac{1}{2}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) + \(\frac{1}{5}\) - ...The alternating behavior helps the series converge, despite the fact that its terms do not go to zero very quickly. This is a fundamental reason why the alternating harmonic series converges conditionally but not absolutely. Here, the alternating nature鈥攐r flipping signs鈥攑lays a crucial role in its behavior and convergence properties.
Absolute Convergence
Absolute convergence is an important concept in understanding series. It refers to a series where the sum of the absolute values of its terms converges.For the alternating harmonic series, if we examine the absolute values of the terms, we get:\[\sum\limits_{n=1}^{\infty} \left|(-1)^{n+1} \frac{1}{n}\right| = \sum\limits_{n=1}^{\infty} \frac{1}{n}\]This series is known as the harmonic series, which diverges. Thus, the alternating harmonic series does not converge absolutely.To visualize this, consider:- If a series converges absolutely, rearranging its terms will lead to the same sum.- Conversely, a conditionally convergent series can have different sums with term rearrangements.Therefore, while the alternating harmonic series converges conditionally, it fails the test for absolute convergence because its absolute series does not converge.
Divergent Series
Divergent series are those that do not have a finite limit. The harmonic series is a classic example of such a series:\[\sum\limits_{n=1}^{\infty} \frac{1}{n}\]Despite appearing simple, it diverges鈥攎eaning it grows without bound. This can be shown using several tests, like the Integral Test or the Comparison Test.Some key points about divergent series:
  • They have no boundary; their terms add up to infinity.
  • Even though terms can get smaller, if they don鈥檛 get smaller fast enough, the series diverges.
This behavior is crucial in determining why the alternating harmonic series is only conditionally convergent.
Alternating Series Test
The Alternating Series Test is a tool used to determine the convergence of series where the terms alternate in sign. This test states that a series:\[\sum\limits_{n=1}^{\infty} (-1)^{n} a_n\]converges if:
  • The absolute value sequence \(a_n\) is non-increasing: \(|a_{n+1}| \leq |a_n|\) for all \(n\).
  • The limit of \(a_n\) as \(n\) approaches infinity is zero: \(\lim\limits_{n\to\infty} a_n = 0\).
Applying this test to the alternating harmonic series:- The sequence of absolute values, \(\frac{1}{n}\), is non-increasing.- It approaches zero as \(n\) gets larger.By meeting these conditions, we confirm that the alternating harmonic series is indeed convergent, specifically conditionally convergent as the absolute version of the series diverges.

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Most popular questions from this chapter

Determine whether the following statements are true and give an explanation or counterexample. a. \(\sum_{k=1}^{\infty}\left(\frac{\pi}{e}\right)^{-k}\) is a convergent geometric series. b. If \(a\) is a real number and \(\sum_{k=12}^{\infty} a^{k}\) converges, then \(\sum_{k=1}^{\infty} a^{k}\) converges. If the series \(\sum_{k=1}^{\infty} a^{k}\) converges and \(|a|<|b|,\) then the series \(\sum_{k=1}^{\infty} b^{k}\) converges. d. Viewed as a function of \(r,\) the series \(1+r^{2}+r^{3}+\cdots\) takes on all values in the interval \(\left(\frac{1}{2}, \infty\right)\) e. Viewed as a function of \(r,\) the series \(\sum_{k=1}^{\infty} r^{k}\) takes on all values in the interval \(\left(-\frac{1}{2}, \infty\right)\)

The Fibonacci sequence \(\\{1,1,2,3,5,8,13, \ldots\\}\) is generated by the recurrence relation \(f_{n+1}=f_{n}+f_{n-1},\) for \(n=1,2,3, \ldots,\) where \(f_{0}=1, f_{1}=1\). a. It can be shown that the sequence of ratios of successive terms of the sequence \(\left\\{\frac{f_{n+1}}{f_{n}}\right\\}\) has a limit \(\varphi .\) Divide both sides of the recurrence relation by \(f_{n},\) take the limit as \(n \rightarrow \infty,\) and show that \(\varphi=\lim _{n \rightarrow \infty} \frac{f_{n+1}}{f_{n}}=\frac{1+\sqrt{5}}{2} \approx 1.618\). b. Show that \(\lim _{n \rightarrow \infty} \frac{f_{n-1}}{f_{n+1}}=1-\frac{1}{\varphi} \approx 0.382\). c. Now consider the harmonic series and group terms as follows: $$\sum_{k=1}^{\infty} \frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\left(\frac{1}{4}+\frac{1}{5}\right)+\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)$$ $$+\left(\frac{1}{9}+\cdots+\frac{1}{13}\right)+\cdots$$ With the Fibonacci sequence in mind, show that $$\sum_{k=1}^{\infty} \frac{1}{k} \geq 1+\frac{1}{2}+\frac{1}{3}+\frac{2}{5}+\frac{3}{8}+\frac{5}{13}+\cdots=1+\sum_{k=1}^{\infty} \frac{f_{k-1}}{f_{k+1}}.$$ d. Use part (b) to conclude that the harmonic series diverges. (Source: The College Mathematics Journal, 43, May 2012)

Consider the following infinite series. a. Write out the first four terms of the sequence of partial sums. b. Estimate the limit of \(\left\\{S_{n}\right\\}\) or state that it does not exist. $$\sum_{k=1}^{\infty}(-1)^{k} k$$

The Greeks solved several calculus problems almost 2000 years before the discovery of calculus. One example is Archimedes' calculation of the area of the region \(R\) bounded by a segment of a parabola, which he did using the "method of exhaustion." As shown in the figure, the idea was to fill \(R\) with an infinite sequence of triangles. Archimedes began with an isosceles triangle inscribed in the parabola, with area \(A_{1}\), and proceeded in stages, with the number of new triangles doubling at each stage. He was able to show (the key to the solution) that at each stage, the area of a new triangle is \(\frac{1}{8}\) of the area of a triangle at the previous stage; for example, \(A_{2}=\frac{1}{8} A_{1},\) and so forth. Show, as Archimedes did, that the area of \(R\) is \(\frac{4}{3}\) times the area of \(A_{1}\).

a. Sketch the function \(f(x)=1 / x\) on the interval \([1, n+1]\) where \(n\) is a positive integer. Use this graph to verify that $$\ln (n+1)<1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}<1+\ln n.$$ b. Let \(S_{n}\) be the sum of the first \(n\) terms of the harmonic series, so part (a) says \(\ln (n+1)0,\) for \(n=1,2,3, \ldots\) c. Using a figure similar to that used in part (a), show that $$\frac{1}{n+1}>\ln (n+2)-\ln (n+1).$$ d. Use parts (a) and (c) to show that \(\left\\{E_{n}\right\\}\) is an increasing sequence \(\left(E_{n+1}>E_{n}\right)\). e. Use part (a) to show that \(\left\\{E_{n}\right\\}\) is bounded above by 1 . f. Conclude from parts (d) and (e) that \(\left\\{E_{n}\right\\}\) has a limit less than or equal to \(1 .\) This limit is known as Euler's constant and is denoted \(\gamma\) (the Greek lowercase gamma). g. By computing terms of \(\left\\{E_{n}\right\\}\), estimate the value of \(\gamma\) and compare it to the value \(\gamma \approx 0.5772 .\) (It has been conjectured that \(\gamma\) is irrational.) h. The preceding arguments show that the sum of the first \(n\) terms of the harmonic series satisfy \(S_{n} \approx 0.5772+\ln (n+1)\) How many terms must be summed for the sum to exceed \(10 ?\)
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