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Chapter 9: Problem 3

Determine whether the alternating series converges; justify your answer. $$\sum_{k=1}^{\infty}(-1)^{k+1} \frac{k+1}{3 k+1}$$

Short Answer

Expert verified
The series does not converge because the terms do not approach zero.

Step by step solution

01

Identify the Series Form

The given series is \(\sum_{k=1}^{\infty}(-1)^{k+1} \frac{k+1}{3k+1}\). This series is an alternating series because it has the factor \((-1)^{k+1}\), which causes the terms to switch signs for each successive value of \(k\).
02

Verify Convergence of the Terms to Zero

For the alternating series test, the sequence of terms without the alternating sign, \(b_k = \frac{k+1}{3k+1}\), must converge to zero as \(k\) approaches infinity. We check the limit: \(\lim_{k \to \infty} \frac{k+1}{3k+1} = \lim_{k \to \infty} \frac{1+1/k}{3+1/k} = \frac{1}{3}\), by dividing the numerator and denominator by \(k\). Since this limit is not zero, the terms do not converge to zero.
03

Conclude on Series Convergence Using the Alternating Series Test

The Alternating Series Test stipulates two conditions: 1) the terms \(b_k\) must eventually decrease in magnitude, and 2) \(b_k\) must approach zero as \(k\to\infty\). While \(b_k = \frac{k+1}{3k+1}\) shows eventual decrease, we found in Step 2 \(b_k\) does not converge to zero. Therefore, the series does not satisfy the second condition of the test.

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Key Concepts

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

Alternating Series Test
The Alternating Series Test helps us determine if a series converges when its terms alternate in sign. It's specifically useful for series in the form of \((-1)^k b_k\). Here are the essential criteria the test requires:

  • First, the absolute value of the terms, which are \(b_k\), must eventually decrease. This means as you keep progressing through the series, each term should be smaller in absolute value than the one before it.
  • Secondly, \(b_k\) needs to approach zero as \(k\) becomes infinitely large.
When these two conditions are met, the series converges. However, if either condition is not satisfied, the series fails the test.

In our specific case, even though \(b_k = \frac{k+1}{3k+1}\) does eventually decrease, it does not approach zero. The test, therefore, indicates that our given series does not converge.
Limit of a Sequence
The limit of a sequence is a fundamental concept in understanding series convergence. When we examine \(b_k = \frac{k+1}{3k+1}\), we calculate its limit as \(k\) approaches infinity.

By simplifying \(b_k\), where both numerator and denominator are divided by \(k\), we get \(\lim_{k \to \infty} \frac{1+1/k}{3+1/k}\).
This simplifies to \(\frac{1}{3}\) as \(k\) grows large. Remember, for the alternating series to converge, this limit needs to be zero.

Since \(\frac{1}{3}\) is not zero, the terms of the sequence do not shrink to insignificance as required for alternating series convergence. Thus, this limit showcases why the series diverges.
Series Divergence
When a series diverges, its sum doesn't settle at a finite number. In the context of alternating series, this occurs when the conditions of the Alternating Series Test aren't met. Specifically, if the terms do not trend towards zero, the series is considered divergent.

In our given example, the series \(\sum_{k=1}^{\infty}(-1)^{k+1} \frac{k+1}{3k+1}\) fails to meet the criteria of the Alternating Series Test. This failure consequently reveals its divergence. The series' terms, instead of shrinking, stabilize around the value \(\frac{1}{3}\), failing to fade into insignificance.

The concept of divergence is crucial because it means that no matter how far you progress, the series' sum will continue to grow indefinitely without approaching a single value.

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Most popular questions from this chapter

Find the first four distinct Taylor polynomials about \(x=x_{0},\) and use a graphing utility to graph the given function and the Taylor polynomials on the same screen. $$\ln (x+1) ; x_{0}=0$$

Find the first four distinct Taylor polynomials about \(x=x_{0},\) and use a graphing utility to graph the given function and the Taylor polynomials on the same screen. $$f(x)=e^{-2 x} ; x_{0}=0$$

Find the Taylor polynomials of orders \(n=0,1,2,3\) and 4 about \(x=x_{0},\) and then find the \(n\) th Taylor polynomial for the function in sigma notation. $$\sin \pi x ; x_{0}=\frac{1}{2}$$

Use Maclaurin series to approximate the integral to three decimal-place accuracy. $$\int_{0}^{1 / 2} \tan ^{-1}\left(2 x^{2}\right) d x$$

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 \(\pi^{4} / 90\).
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