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State whether each of the following series converges absolutely, conditionally, or not at all. $$\sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{n+3}$$

Short Answer

Expert verified
The series diverges.

Step by step solution

01

Identify the Series Type

The given series is \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{n+3} \). It alternates in sign due to \((-1)^{n+1}\), which suggests that it is an alternating series.
02

Analyze Convergence Using the Alternating Series Test

For the alternating series \( \sum (-1)^{n+1} a_n \) to converge, \( a_n \) must be decreasing and approach zero as \( n \to \infty \). Let \( a_n = \frac{n}{n+3} \). As \( n \to \infty \), \( a_n \to 1 \), which is not zero. Therefore, the series does not satisfy the conditions of the alternating series test and does not converge.
03

Check Absolute Convergence

To check for absolute convergence, consider the series of the absolute values: \( \sum_{n=1}^{\infty} \left| (-1)^{n+1} \frac{n}{n+3} \right| = \sum_{n=1}^{\infty} \frac{n}{n+3} \). To determine convergence, compare it to the harmonic series \( \sum_{n=1}^{\infty} \frac{n}{n} = \sum_{n=1}^{\infty} 1 \), which diverges. Since for large \(n\), \( \frac{n}{n+3} \approx \frac{n}{n} = 1\), the series \( \sum_{n=1}^{\infty} \frac{n}{n+3} \) also diverges.
04

Conclusion on Convergence

Since the series does not meet the criteria for the alternating series test and does not converge absolutely, it neither converges absolutely nor conditionally.

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Key Concepts

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

Absolute Convergence
Absolute convergence means that if we take the absolute value of each term in a series and the resulting series converges, then the original series also converges. It is a stronger form of convergence which applies to both alternating and non-alternating series.
To determine absolute convergence, we examine the series of absolute values. For the given series, it's \( \sum_{n=1}^{\infty} \left| (-1)^{n+1} \frac{n}{n+3} \right| = \sum_{n=1}^{\infty} \frac{n}{n+3} \).
This resembles the harmonic series \( \sum_{n=1}^{\infty} \frac{1}{n} \), known to diverge. Since this absolute value series diverges, the original series does not converge absolutely.
In essence, if a series does not converge absolutely, it might still converge conditionally. However, this is not the case here as we’ll explore in the following sections.
Conditional Convergence
Conditional convergence occurs when a series converges, but does not converge absolutely. This typically happens in alternating series, where even though the absolute series diverges, the original series might still converge due to the alternating nature of its terms.
However, for a series to be conditionally convergent, it must pass the alternating series test which we'll detail later. Our series \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{n+3} \) fails this test. This means it neither converges absolutely nor conditionally.
Thus, for our series, neither condition holds, so we conclude it diverges completely.
Alternating Series Test
The Alternating Series Test is a handy tool for checking the convergence of series with alternating signs. For a given series \( \sum (-1)^n a_n \), this test stipulates two conditions for convergence:
  • Each subsequent term \( a_n \) must be smaller than the previous term (i.e., \( a_n \) must be decreasing).
  • The limit of \( a_n \) as \( n \) approaches infinity must be zero.
Our series is \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{n+3} \) with \( a_n = \frac{n}{n+3} \). Although it alternates in sign, as \( n \to \infty \), \( a_n \to 1 \), not zero, so it does not satisfy the necessary conditions.
Since the alternating series test fails, the series does not converge via this method, further supporting our findings on divergence.
Divergence
Divergence simply means that the sum of an infinite series does not approach a finite limit. In the case of the series \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{n+3} \), we found that it neither converges absolutely nor conditionally.
Because both the alternating series test and the series of absolute values indicate divergence, we conclude that the overall series diverges.
Divergence should not be viewed negatively but rather as a result of specific criteria not being met. For this series, the failure arises because its terms do not approach zero and the absolute series diverges.

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Most popular questions from this chapter

In each of the following problems, use the estimate \(\left|R_{N}\right| \leq b_{N+1}\) to find a value of \(N\) that guarantees that the sum of the first \(N\) terms of the alternating series \(\sum_{n=1}^{\infty}(-1)^{n+1} b_{n}\) differs from the infinite sum by at most the given error. Calculate the partial sum \(S_{N}\) for this \(N\). $$\text { [T] } b_{n}=1 / \ln (n), \quad n \geq 2, \text { error }<10^{-1}$$

In the following exercises, use either the ratio test or the root test as appropriate to determine whether the series \(\sum_{k=1}^{\infty} a_{k}\) with given terms \(a_{k}\) converges, or state if the test is inconclusive. $$a_{n}=\left(n^{1 / n}-1\right)^{n}$$

[T] Plot the series \(\sum_{n=1}^{100} \frac{\cos (2 \pi n x)}{n^{2}}\) for \(0 \leq x<1\) and describe its graph.

Use the ratio test to determine whether \(\sum_{n=1}^{\infty} a_{n}\) converges, where \(a_{n}\) is given in the following problems. State if the ratio test is inconclusive. $$\sum_{n=1}^{\infty} \frac{(2 n) !}{n^{2 n}}$$

Let \(a_{n}=2^{-[n / 2]}\) where \([x]\) is the greatest integer less than or equal to \(x\). Determine whether \(\sum_{n=1}^{\infty} a_{n}\) converges and justify your answer. The following advanced exercises use a generalized ratio test to determine convergence of some series that arise in particular applications when tests in this chapter, including the ratio and root test, are not powerful enough to determine their convergence. The test states that if \(\lim _{n \rightarrow \infty} \frac{a_{2 n}}{a_{n}}<1 / 2\), then \(\sum a_{n}\) converges, while if \(\lim _{n \rightarrow \infty} \frac{a_{2 n+1}}{a_{n}}>1 / 2\), then \(\sum a_{n}\) diverges.

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