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Classify each series as absolutely convergent, conditionally convergent, or divergent. $$\sum_{k=1}^{\infty} \frac{\cos k \pi}{k}$$

Short Answer

Expert verified
The series is conditionally convergent.

Step by step solution

01

Identify Series Type

The given series is \( \sum_{k=1}^{\infty} \frac{\cos k \pi}{k} \). Note that \( \cos k \pi \) alternates between -1 and 1 as \( k \) increases. Therefore, the series can be rewritten using the alternating series notation as \( \sum_{k=1}^{\infty} \frac{(-1)^k}{k} \). This is an alternating harmonic series.
02

Apply Alternating Series Test

The alternating series test states that a series \( \sum_{k=1}^{\infty} (-1)^k a_k \) converges if \( a_k \) is positive, decreasing, and approaches zero as \( k \rightarrow \infty \). In our case, \( a_k = \frac{1}{k} \) is positive and decreases to zero as \( k \rightarrow \infty \). Hence, the series \( \sum_{k=1}^{\infty} \frac{(-1)^k}{k} \) converges.
03

Check for Absolute Convergence

To check for absolute convergence, consider the absolute value of the series terms: \( \sum_{k=1}^{\infty} \left|\frac{(-1)^k}{k}\right| = \sum_{k=1}^{\infty} \frac{1}{k} \). This becomes the harmonic series, which is known to diverge. Hence, the series does not converge absolutely.
04

Classify the Series

Since the series converges due to the alternating series test but does not converge absolutely, it is classified as conditionally convergent.

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Key Concepts

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

Alternating Series Test
An alternating series is a series where the terms alternate in sign. A classic example involves terms like \((-1)^k\), where each term has opposite sign to its predecessor. The Alternating Series Test is a tool we use to determine if this type of series converges. According to the test, for a series \( \sum_{k=1}^{\infty} (-1)^k a_k \) to converge, two main conditions need to be satisfied:
  • The absolute value of the terms \(a_k\) must be decreasing. This means each term should be smaller in magnitude than the one before it: \(a_{k+1} \leq a_k\).

  • The limit of \(a_k\) as \(k\) approaches infinity should be zero: \( \lim_{k \to \infty} a_k = 0 \).
In the given series, \( a_k = \frac{1}{k} \), this function meets both criteria: it is decreasing since each subsequent term is smaller as \(k\) increases, and as \(k\) approaches infinity, \(a_k\) approaches zero. Therefore, using the Alternating Series Test, we conclude that this series converges.
Absolute Convergence
To understand absolute convergence, consider the series with absolute terms. A series is said to converge absolutely if the series formed by taking the absolute value of each term is itself convergent. This is a stronger form of convergence since if a series converges absolutely, it also converges.
Considering our series \( \sum_{k=1}^{\infty} \left|\frac{(-1)^k}{k}\right| \), which simplifies to the harmonic series \(\sum_{k=1}^{\infty} \frac{1}{k} \). Unfortunately, the harmonic series is a well-known divergent series. This lack of convergence in absolute terms means that our original series does not converge absolutely.
When analyzing a series, finding that it does not converge absolutely doesn't automatically tell us about overall convergence. That is why we must explore conditional convergence, as we'll see next.
Conditional Convergence
Conditional convergence is an interesting phenomenon. A series is conditionally convergent if it converges, but does not converge absolutely. This means while the series itself adds up to a finite number, taking absolute values of all its terms causes the series to diverge.
  • This type of convergence often arises in alternating series, just like our example \( \sum_{k=1}^{\infty} \frac{(-1)^k}{k} \).
  • The series converges due to its alternating nature (as verified by the Alternating Series Test), but fails the test for absolute convergence due to the divergently behaving harmonic series.
Hence, our original series is conditionally convergent. Although it might not seem intuitive at first, conditional convergence is a useful concept in mathematical analysis, especially in understanding the behavior of complex series and functions.

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Most popular questions from this chapter

Determine whether the series converges. $$\sum_{k=1}^{\infty}\left(1+\frac{1}{k}\right)^{-k}$$

(a) Show that the hypotheses of the integral test are satisfied by the series \(\sum_{k=1}^{\infty} 1 /\left(k^{3}+1\right)\) (b) Use a CAS and the integral test to confirm that the series converges. (c) Construct a table of partial sums for \(n=10,20\) \(30, \ldots, 100,\) showing at least six decimal places. (d) Based on your table, make a conjecture about the sum of the series to three decimal-place accuracy. (e) Use part (b) of Exercise 36 to check your conjecture.

In each part, find all values of \(x\) for which the series converges, and find the sum of the series for those values of \(x\) (a) \(x-x^{3}+x^{5}-x^{7}+x^{9}-\cdots\) (b) \(\frac{1}{x^{2}}+\frac{2}{x^{3}}+\frac{4}{x^{4}}+\frac{8}{x^{5}}+\frac{16}{x^{6}}+\cdots\) (c) \(e^{-x}+e^{-2 x}+e^{-3 x}+e^{-4 x}+e^{-5 x}+\cdots\)

Exercise will show how a partial sum can be used to obtain upper and lower bounds on the sum of a series when the hypotheses of the integral test are satisfied. This result will be needed in Exercises. It was stated in Exercise 35 that $$\sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90}$$ (a) Let \(s_{n}\) be the \(n\) th partial sum of the series above. Show that $$ s_{n}+\frac{1}{3(n+1)^{3}} < \frac{\pi^{4}}{90} < s_{n}+\frac{1}{3 n^{3}} $$ (b) We can use a partial sum of the series to approximate \(\pi^{4} / 90\) to three decimal-place accuracy by capturing the sum of the series in an interval of length 0.001 (or less). Find the smallest value of \(n\) such that the interval containing \(\pi^{4} / 90\) in part (a) has a length of 0.001 or less. (c) Approximate \(\pi^{4} / 90\) to three decimal places using the midpoint of an interval of width at most 0.001 that contains the sum of the series. Use a calculating utility to confirm that your answer is within 0.0005 of \(\pi^{4} / 90\).

Writing The sine function can be defined geometrically from the unit circle or analytically from its Maclaurin series. Discuss the advantages of each representation with regard to providing information about the sine function.

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