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\(19-40=\) Determine whether the series is absolutely convergent, conditionally convergent, or divergent. $$ \sum_{n=1}^{\infty}(-1)^{n-1} \frac{n}{n^{2}+4} $$

Short Answer

Expert verified
The series is conditionally convergent.

Step by step solution

01

Recall the Definitions

An infinite series \(\sum a_n\) is called **absolutely convergent** if the series of absolute values \(\sum |a_n|\) is convergent. It is called **conditionally convergent** if \(\sum a_n\) is convergent, but \(\sum |a_n|\) is divergent. If \(\sum a_n\) is divergent, we call the series **divergent**.
02

Identify the Series Terms

The given series is \(\sum_{n=1}^{\infty} (-1)^{n-1} \frac{n}{n^2+4}\), which is an alternating series with terms \(a_n = (-1)^{n-1} \frac{n}{n^2+4}\).
03

Apply the Alternating Series Test

An alternating series \(\sum (-1)^{n} a_n\) converges if the sequence \(a_n\) is positive, decreasing, and \(\lim_{n \to \infty} a_n = 0\). In this series, \(a_n = \frac{n}{n^2+4}\) is positive for all \(n\). Let's check the limit. \[ \lim_{n \to \infty} \frac{n}{n^2 + 4} = \lim_{n \to \infty} \frac{1}{n + \frac{4}{n}} = 0. \]This means the series passes the second condition of the Alternating Series Test.
04

Check Decreasing Nature

To check if \(a_n = \frac{n}{n^2+4}\) is decreasing, let's compare \(a_n\) and \(a_{n+1}\). We have:\[a_{n+1} = \frac{n+1}{(n+1)^2 + 4} = \frac{n+1}{n^2 + 2n + 5}.\]By comparing \(\frac{n}{n^2+4}\) and \(\frac{n+1}{n^2+2n+5}\), or by checking the derivative, we find \( rac{n}{n^2+4} > \frac{n+1}{n^2+2n+5}\) for all \(n\geq 1\). Thus, \(a_n\) is decreasing.
05

Conclusion on Conditional Convergence

Since the alternating series converges and we have confirmed that \(\lim_{n \to \infty} a_n = 0\) and \(a_n\) is decreasing, the original series is conditionally convergent.
06

Apply Absolute Convergence Test

Check the series of the absolute values: \(\sum \left|\frac{n}{n^2+4}\right|\). Notice that \(\frac{n}{n^2+4}\) is greater than \(\frac{1}{n}\) for large \(n\), indicating that \(\sum \frac{n}{n^2+4}\) is divergent because the series \(\sum \frac{1}{n}\) (harmonic series) is divergent.
07

Conclusion on Absolute Convergence

Since the series of absolute values diverges, the original series is not absolutely convergent.

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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 is crucial for determining whether an alternating series converges. An alternating series takes the form \( \sum_{n=1}^{\infty} (-1)^n a_n \), where the terms \( a_n \) can change signs with each term. The test provides three conditions to confirm convergence:

  • Each term \( a_n \) must be positive.
  • \( a_n \) must form a decreasing sequence.
  • The sequence must approach zero as \( n \) approaches infinity, i.e., \( \lim_{n \to \infty} a_n = 0 \).
For the series \( \sum_{n=1}^{\infty} (-1)^{n-1} \frac{n}{n^2+4} \), we identified that:

- Each term \( \frac{n}{n^2+4} \) is positive.
- The sequence is decreasing as shown by calculations and derivative checks.
- The limit of \( \frac{n}{n^2+4} \) goes to zero, satisfying all conditions. Therefore, the series converges by the Alternating Series Test.
Absolute Convergence
Absolute convergence of a series \( \sum a_n \) means the series \( \sum |a_n| \) also converges. If a series is absolutely convergent, it implies the series converges regardless of the order of its terms. To determine absolute convergence for \( \sum_{n=1}^{\infty} (-1)^{n-1} \frac{n}{n^2+4} \), we check \( \sum \left|\frac{n}{n^2+4}\right| \).

The comparison to the harmonic series \( \sum \frac{1}{n} \) shows that \( \frac{n}{n^2+4} > \frac{1}{n} \) for large \( n \). Since the harmonic series diverges, so does \( \sum \left|\frac{n}{n^2+4}\right| \), meaning the original series is not absolutely convergent.
Conditional Convergence
Conditional convergence occurs when an alternating series converges, but the absolute series does not. This is exactly what we have with \( \sum_{n=1}^{\infty} (-1)^{n-1} \frac{n}{n^2+4} \).

While the series was shown to converge via the Alternating Series Test, the absolute series diverges because it resembles a divergent harmonic series. Thus, though the original series converges, it depends on its alternating nature to do so. This makes the series conditionally convergent.

For practical purposes, conditional convergence implies caution, as rearranging terms could potentially change the outcome if not careful. However, within the context of alternating series, its convergence remains as long as the order is maintained.
Divergent Series
A divergent series fails to meet the criteria for either absolute or conditional convergence. In simpler terms, neither the original series nor its series of absolute values converges.

With our initial series \( \sum_{n=1}^{\infty} (-1)^{n-1} \frac{n}{n^2+4} \), while it is not absolutely convergent, it still converges conditionally. This takes it out of the diverging category. However, had both checks fallen through, we would label it as divergent.

Divergent series, such as the classic \( \sum \frac{1}{n} \) or geometric series with a ratio \( |r| \ge 1 \), are essential concepts that illustrate when terms of a series grow without bound or oscillate erratically, leading to no single sum value.

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Most popular questions from this chapter

Find the sum of the series. $$\sum_{n=0}^{\infty} \frac{(-1)^{n} \pi^{2 n}}{6^{2 n}(2 n) !}$$

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If the \(n\) th partial sum of a series \(\sum_{n=1}^{\infty} a_{n}\) is $$s_{n}=\frac{n-1}{n+1}$$ find \(a_{n}\) and \(\sum_{n=1}^{\infty} a_{n}\)

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