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(a) Find examples to show that if \(\sum a_{k}\) converges, then \(\sum a_{k}^{2}\) may diverge or converge. (b) Find examples to show that if \(\sum a_{k}^{2}\) converges, then \(\sum a_{k}\) may diverge or converge.

Short Answer

Expert verified
\(a_k = \frac{(-1)^{k-1}}{\sqrt{k}}\) shows if \(\sum a_k\) converges, \(\sum a_k^2\) diverges; if \(\sum a_k^2\) converges, \(\sum a_k\) diverges. \(a_k = \frac{1}{k^2}\) shows both \(\sum a_k\) and \(\sum a_k^2\) can converge.

Step by step solution

01

Understanding the Problem

We need to find examples for both parts of the problem statement: (a) where the series \(\sum a_k\) converges but \(\sum a_k^2\) may diverge or converge, and (b) where \(\sum a_k^2\) converges but \(\sum a_k\) may diverge or converge.
02

Example for Part (a) - Convergent \(\sum a_k\) and Divergent \(\sum a_k^2\)

Consider \(a_k = \frac{1}{k}\). The harmonic series \(\sum \frac{1}{k}\) diverges, but if the \(a_k\) were to converge, \(a_k\) would need to be a different sequence. Let's modify it: consider \(a_k = \frac{(-1)^{k-1}}{\sqrt{k}}\). The series \(\sum \frac{(-1)^{k-1}}{\sqrt{k}}\) converges by the Alternating Series Test, however, \(\sum a_k^2 = \sum \frac{1}{k}\), which diverges.
03

Example for Part (a) - Convergent \(\sum a_k\) and Convergent \(\sum a_k^2\)

Consider \(a_k = \frac{1}{k^2}\). The series \(\sum \frac{1}{k^2}\) converges since it is a p-series with \(p=2 > 1\). Also the series \(\sum a_k^2 = \sum \frac{1}{k^4}\) converges as another p-series with \(p=4 > 1\).
04

Example for Part (b) - Convergent \(\sum a_k^2\) and Divergent \(\sum a_k\)

Use the same example considered earlier for divergence: \(a_k = \frac{(-1)^{k-1}}{\sqrt{k}}\). Here, \(\sum a_k^2 = \sum \frac{1}{k}\) converges, but \(\sum a_k\) diverges even if \(\sum a_k^2\) converges.
05

Example for Part (b) - Convergent \(\sum a_k^2\) and Convergent \(\sum a_k\)

Again using \(a_k = \frac{1}{k^2}\), \(\sum a_k\) converges and so does \(\sum a_k^2\) as previously described in the Part (a) examples.

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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 a helpful tool in determining the convergence of series whose terms alternate in sign, like positive and negative. To use this test, the series must meet two conditions.
  • The absolute value of the terms must decrease monotonically. This means, if you have terms like: \(a_1, a_2, a_3, \ldots\), for the series to pass the test, you need \(|a_{k+1}| \leq |a_k|\) for all k.
  • The limit of the terms must be zero as k approaches infinity: \(\lim_{{k \to \infty}} a_k = 0\).
Let’s see this in practice. For example, with the series \(\sum \frac{(-1)^{k-1}}{\sqrt{k}}\), the terms alternate between positive and negative. Each term becomes smaller since \(\frac{1}{\sqrt{k+1}}\) is always smaller than \(\frac{1}{\sqrt{k}}\). Also, as k increases, the terms tend towards zero. Thus, this series converges as per the Alternating Series Test.
p-series
A p-series is a series of the form \(\sum \frac{1}{k^p}\), where \(p\) is a constant value. Not every p-series converges. The rule here depends on the exponent \(p\):
  • If \(p > 1\), the p-series converges. This is because the terms diminish quickly enough to reach convergence.
  • If \(p \leq 1\), the series diverges. In this case, the terms do not diminish quickly enough, leading the series to diverge.
For example, consider the series \(\sum \frac{1}{k^2}\). Here, \(p = 2 > 1\), so this series converges. Furthermore, if you look at \(\sum \frac{1}{k^4}\), it's another p-series where \(p = 4 > 1\), ensuring convergence.
Harmonic Series
The harmonic series is one of the most classic examples of divergence, and it is written as \(\sum \frac{1}{k}\). At first glance, the terms get smaller and somewhat suggest convergence. However, the series diverges.
For a better understanding, consider the harmonic series as a specific case of a p-series with \(p = 1\). Since \(p = 1\) falls into the \(p \leq 1\) category, it diverges.
An interesting comparison is with the alternating harmonic series \(\sum \frac{(-1)^{k-1}}{k}\). Despite being based on the same terms \(\frac{1}{k}\), the alternating feature allows it to pass the Alternating Series Test, hence it converges. This highlights how small changes in a series can significantly affect its convergence properties.

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