Problem 29 - Calculus: One and Several Variables

Question

Test these series for (a) absolute convergence, (b) conditional convergence. \(\sum \frac{\sin (\pi k / 4)}{k^{2}}\).

Step-by-Step Solution

Answer

The series \(\sum \frac{\sin (\pi k / 4)}{k^2}\) is absolutely convergent, as we proved using the Comparison test with \(\frac{1}{k^2}\), a convergent p-series. Since it is absolutely convergent, it is also convergent. Conditional convergence does not occur in this case because an absolutely convergent series cannot be conditionally convergent.

Step 1: Test for absolute convergence

To test for absolute convergence, we will study the convergence of the absolute value of the series: \(\sum \left|\frac{\sin (\pi k / 4)}{k^2}\right|\) We want to investigate whether the series converges or not. To do that, we will use the Comparison test.

Step 2: Apply the Comparison test

For the Comparison test, we need to find a known convergent or divergent series of non-negative terms to compare with our given series. Since \(\left|\sin (\pi k / 4)\right| \leq 1\), we can use \(\frac{1}{k^2}\) as a comparison, which is a convergent p-series when p=2. Thus, we have: \(0 \leq \left|\frac{\sin (\pi k / 4)}{k^2}\right| \leq \frac{1}{k^2} \) Now, we know that \(\sum \frac{1}{k^2}\) converges by the p-series test. By the Comparison test, the series \(\sum \left|\frac{\sin (\pi k / 4)}{k^2}\right|\) should also converge. Since the absolute value of the series converges, the original series \(\sum \frac{\sin (\pi k / 4)}{k^2}\) is absolutely convergent.

Step 3: Test for conditional convergence

To test for conditional convergence, we check whether the original series converges when the absolutely convergent series does not. However, in our case, the original series is already absolutely convergent, which means conditional convergence cannot happen.

Step 4: Conclusion

The series \(\sum \frac{\sin (\pi k / 4)}{k^2}\) is absolutely convergent, which implies that it is also convergent. It does not exhibit conditional convergence because an absolutely convergent series cannot be conditionally convergent.

Key Concepts

Absolute Convergence

When we talk about absolute convergence, we are checking if a series still converges when we take the absolute value of every term. This is a stronger form of convergence. In mathematical terms, a series \( \sum a_k \) is said to be absolutely convergent if the series of absolute values \( \sum |a_k| \) converges.

The important part here is that absolute convergence guarantees regular convergence too. This means if the absolute values add up to a finite number, the original series does as well.

For instance, if applying the absolute value of the series results in a convergent p-series, or any known convergent series, then we conclude our original series is absolutely convergent. In the exercise, the series \( \sum \left|\frac{\sin (\pi k / 4)}{k^2}\right| \) is shown to converge using the comparison test with the p-series \( \sum \frac{1}{k^2} \). This leads to the whole series being absolutely convergent, which simplifies our understanding of convergence considerably.

Conditional Convergence

Conditional convergence is a bit trickier to grasp, but here’s the idea: a series is conditionally convergent if it converges, but does not converge absolutely. This means even though the series converges, the series of absolute terms \( \sum |a_k| \) diverges.

To illustrate, think of conditional convergence as a sort of balance. The terms in the series may swing positive and negative, allowing them to 'cancel' each other out enough to converge. A famous example is the alternating harmonic series, which converges despite the harmonic series itself being divergent.

In the exercise, once we ascertain absolute convergence, checking for conditional convergence becomes unnecessary. That's because a series cannot simultaneously be absolutely and conditionally convergent. The exercise confirms that the original series \( \sum \frac{\sin (\pi k / 4)}{k^2} \) is absolutely convergent, ruling out the possibility of conditional convergence.

Comparison Test

The Comparison Test is a handy tool when it comes to proving convergence of series. It lets you compare a given series with another known series to determine convergence or divergence. The idea is simple: you pick a series that's easy to deal with and compare term-by-term.

There are two parts to this test:

• If \( 0 \leq a_k \leq b_k \) for all \( k \), and \( \sum b_k \) converges, then \( \sum a_k \) converges too.
• If \( 0 \leq b_k \leq a_k \) for all \( k \), and \( \sum b_k \) diverges, then \( \sum a_k \) diverges too.

In our series check, we compared our terms \( \left|\frac{\sin (\pi k / 4)}{k^2}\right| \) to the simpler and well-known convergent series \( \sum \frac{1}{k^2} \). Since \( \left|\sin (\pi k / 4)\right| \leq 1 \) ensures \( 0 \leq \left|\frac{\sin (\pi k / 4)}{k^2}\right| \leq \frac{1}{k^2} \), the comparison confirmed the convergence of the more complex series.

P-Series Test

The p-series test involves a series of the form \( \sum \frac{1}{k^p} \). Here, the value of \( p \) determines whether the series converges or diverges.

The rule is straightforward:

• If \( p > 1 \), the series \( \sum \frac{1}{k^p} \) converges.
• If \( p \leq 1 \), the series \( \sum \frac{1}{k^p} \) diverges.

Understanding this test is crucial because it often serves as a reference point in the Comparison Test.

In contexts like our exercise, we use the fact that \( \sum \frac{1}{k^2} \) is convergent since \( p = 2 > 1 \). It acts as a reliable benchmark series. This makes it easier to assert that the original series is absolutely convergent, simplifying the analysis of more complex series.