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Classify each series as absolutely convergent, conditionally convergent, or divergent. $$ \sum_{k=1}^{\infty}(-1)^{k+1} \frac{3^{2 k-1}}{k^{2}+1} $$

Short Answer

Expert verified
The series is divergent.

Step by step solution

01

Recognize the Series Type

The given series is \( \sum_{k=1}^{\infty}(-1)^{k+1} \frac{3^{2 k-1}}{k^{2}+1} \), which is an alternating series because of the \((-1)^{k+1}\) term.
02

Test for Absoluteness

To determine if the given series is absolutely convergent, consider the absolute value of the series: \( \sum_{k=1}^{\infty} \left|(-1)^{k+1} \frac{3^{2 k-1}}{k^{2}+1}\right| = \sum_{k=1}^{\infty} \frac{3^{2k-1}}{k^{2}+1} \). We need to check whether this series converges.
03

Check Convergence of Absolute Series

For \( \sum_{k=1}^{\infty} \frac{3^{2k-1}}{k^{2}+1} \), compare it to the series \( \sum_{k=1}^{\infty} \frac{3^{2k-1}}{k^2}\). It can be shown that \( \frac{3^{2k-1}}{k^{2}+1} > \frac{3^{2k-1}}{k^2} \). The series \( \frac{3^{2k-1}}{k^2} \) diverges (as a geometric series with common ratio \(r = 9\) where \(r > 1\)). Hence, \( \sum_{k=1}^{\infty} \frac{3^{2k-1}}{k^{2}+1} \) also diverges.
04

Apply the Alternating Series Test

Apply the alternating series test to the original series \( \sum_{k=1}^{\infty}(-1)^{k+1} \frac{3^{2 k-1}}{k^{2}+1} \). The terms do not tend to zero due to the exponential growth in the numerator. Therefore, the alternating series test fails.
05

Classify the Series

Since the absolute series diverges and the alternating series test fails, \( \sum_{k=1}^{\infty}(-1)^{k+1} \frac{3^{2 k-1}}{k^{2}+1} \) is divergent, not absolutely or 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
When tackling series convergence problems, understanding the alternating series test is essential. An alternating series is identified by terms that switch signs, typically seen in the format \[(-1)^{k+1} a_k\] This back-and-forth pattern is what makes a series "alternating." The alternating series test helps determine whether such series converge. It basically requires two conditions:
  • The absolute value of the terms \( a_k \) must be decreasing
  • The limit of \( a_k \) as \( k \) approaches infinity must be zero
If both conditions are met, we can safely say the alternating series is convergent. However, in our example, the series \(\sum_{k=1}^{\infty}(-1)^{k+1} \frac{3^{2 k-1}}{k^{2}+1}\) fails the test because the terms do not shrink to zero. This result indicates that the series does not converge by the alternating series test.
Absolute Convergence
Absolute convergence is a stronger form of convergence than conditional convergence. A series is absolutely convergent if the series of absolute values is convergent. In simpler terms, it means that the series \(\sum_{k=1}^{\infty}|a_k|\) converges. If a series is absolutely convergent, it ensures the series itself is convergent regardless of the arrangement of terms. To analyze for absolute convergence, we take the absolute form of the series and check its convergence status. For our example, when we examine \(\sum_{k=1}^{\infty} \frac{3^{2k-1}}{k^{2}+1}\), we find it diverges. This is done by comparing it to another series, such as \(\sum_{k=1}^{\infty} \frac{3^{2k-1}}{k^2}\). The comparison tells us that the absolute series diverges, so the original series is not absolutely convergent.
Conditional Convergence
Conditional convergence occurs when a series converges, but its absolute counterpart does not converge. It reflects a delicate balance where the alternating nature of the terms allows for convergence even if their magnitudes (absolute values) would lead to divergence. A concept often involved here is the harmonic series and its various extensions. For a series to be conditionally convergent, it must pass the alternating series test but fail the absolute convergence test. When only the former is successful, the series remains conditionally convergent. However, in our scenario with \(\sum_{k=1}^{\infty}(-1)^{k+1} \frac{3^{2k-1}}{k^{2}+1}\), both convergence tests fail. The terms do not taper to zero, failing the alternating series test, and the absolute series diverges, precluding absolute convergence. Therefore, this series is not conditionally convergent.

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Most popular questions from this chapter

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

The limit of an indeterminate form as \(x \rightarrow x_{0}\) can sometimes be found by expanding the functions involved in Taylor series about \(x=x_{0}\) and taking the limit of the series term by term. Use this method to find the limits in these exercises. $$ \text { (a) } \lim _{x \rightarrow 0} \frac{1-\cos x}{\sin x} \quad \text { (b) } \lim _{x \rightarrow 0} \frac{\ln \sqrt{1+x}-\sin 2 x}{x} $$

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