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Chapter 9: Problem 28

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 conditionally convergent.

Step by step solution

01

Determine Absolute Convergence

To check for absolute convergence, we consider the series without the alternating sign: \[ \sum_{k=1}^{\infty} \frac{3^{2k-1}}{k^2+1} \]To check convergence, use the Comparison Test or the Ratio Test.
02

Apply Comparison Test

Compare \( \frac{3^{2k-1}}{k^2+1} \) with \( \frac{3^{2k}}{k^2} \). Since \( 3^{2k-1} = \frac{3^{2k}}{3} \), we have:\[ \frac{3^{2k-1}}{k^2+1} < \frac{3^{2k}}{k^2} \]The series \( \sum_{k=1}^{\infty} \frac{3^{2k}}{k^2} \) diverges because \( 3^{2k} \) grows exponentially.
03

Examine Conditional Convergence Using Alternating Series Test

Consider the original alternating series: \[ \sum_{k=1}^{\infty} (-1)^{k+1} \frac{3^{2k-1}}{k^2+1} \]The Alternating Series Test states that a series \( \sum (-1)^k a_k \) converges if \( a_k > 0 \), \( a_k \to 0 \), and \( a_{k+1} \leq a_k \). Here, \( a_k = \frac{3^{2k-1}}{k^2+1} \). Check if \( a_k \to 0 \) and \( a_{k+1} \leq a_k \).
04

Check Alternating Series Test Conditions

As \( k \to \infty \), \( \frac{3^{2k-1}}{k^2+1} \to 0 \). Next, check if \( \frac{3^{2k+1}}{(k+1)^2+1} \leq \frac{3^{2k-1}}{k^2+1} \). Since \( 3^{2k} \) grows faster than polynomial terms in the denominator, \( \frac{3^{2k+1}}{(k+1)^2+1} \leq \frac{3^{2k-1}}{k^2+1} \) holds for large \( k \). Thus, the alternating series converges.

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Key Concepts

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

Comparison Test
The Comparison Test is a useful tool when determining whether a series converges or diverges. The idea is simple: you compare the series you are interested in to another series that you already know the convergence behavior of.
  • If the series you are testing is smaller term-by-term than a known convergent series, then your series converges.
  • If your series is larger term-by-term than a divergent series, then your series diverges.
In the given exercise, the series \[ \sum_{k=1}^{\infty} \frac{3^{2k-1}}{k^2+1} \]is compared to \[ \sum_{k=1}^{\infty} \frac{3^{2k}}{k^2} \]. Since \( 3^{2k} \) grows exponentially鈥攁 rate faster than any polynomial in the denominator鈥攖he series diverges. This comparison helps us understand the divergence of the original series as well.
Ratio Test
The Ratio Test helps determine the absolute convergence of a series by examining the limit of the ratio of successive terms. The test is conducted as follows:If you have a series \( \sum a_k \),you calculate\(\lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right|.\)
  • If this limit is less than 1, the series converges absolutely.
  • If the limit is greater than 1, the series diverges.
  • If the limit equals 1, the test is inconclusive.
In the problem provided, they opted for the comparison test over this test. However, the Ratio Test could be another way to analyze if the series without the alternating component converges absolutely by looking at the exponential growth of \( 3^{2k} \).
Alternating Series Test
The Alternating Series Test is specifically designed for series where terms alternate in sign. A classic example is series like:\[ \sum(-1)^k a_k \].Here, you apply the test by ensuring two main conditions:
  • The absolute value of the terms \( a_k \) decreases monotonically (i.e., \( a_{k+1} \leq a_k \)).
  • The terms \( a_k \) should approach zero as \( k \)goes to infinity.
In this exercise, the series \[ \sum_{k=1}^{\infty} (-1)^{k+1} \frac{3^{2k-1}}{k^2+1} \] satisfies both conditions:
  • The terms \( a_k = \frac{3^{2k-1}}{k^2+1} \) approach zero because the numerator grows slower than the denominator as \( k \to \infty \).
  • The sequence \( a_k \)decreases since, beyond a certain point, exponential growth in the numerator is counteracted by quadratic growth in the denominator, meeting the monotonicity requirement.
Thus, by the Alternating Series Test, the series converges conditionally.

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Most popular questions from this chapter

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