Problem 38
Question
a. Show that if \(\sum a_{n}\) converges absolutely, then \(\sum a_{n}^{2}\) converges. b. Show that the converse of the result in part (a) is false by finding a series \(\sum a_{n}\) for which \(\sum a_{n}^{2}\) converges, but \(\Sigma\left|a_{n}\right|\) diverges.
Step-by-Step Solution
Verified Answer
(a) If \(\sum a_{n}\) converges absolutely, we have \(0 \leq a_{n}^{2} \leq |a_{n}|^{2}\) for all terms. By the comparison test, since \(\sum |a_{n}|\) converges, then \(\sum a_{n}^{2}\) must also converge.
(b) The series \(\sum a_{n}\) with \(a_n = (-1)^{n}\frac{1}{\sqrt{n}}\) disproves the converse. The sum of squares \(\sum a_{n}^{2}\) converges (since it is a harmonic series), while the sum of absolute values \(\sum |a_{n}|\) diverges (since it is a p-series with \(p = \frac{1}{2}\)).
1Step 1: Part (a)
If \(\sum |a_n|\) converges, then \(|a_n| \to 0\), so for large \(n\), \(|a_n| \leq 1\), hence \(a_n^2 = |a_n|^2 \leq |a_n|\).
By the Comparison Test, since \(\sum |a_n|\) converges and \(0 \leq a_n^2 \leq |a_n|\), the series \(\sum a_n^2\) converges.
By the Comparison Test, since \(\sum |a_n|\) converges and \(0 \leq a_n^2 \leq |a_n|\), the series \(\sum a_n^2\) converges.
2Step 2: Part (b) - Counterexample
Let \(a_n = \frac{1}{n}\). Then \(\sum a_n^2 = \sum \frac{1}{n^2}\) converges (p-series with \(p = 2 > 1\)).
But \(\sum |a_n| = \sum \frac{1}{n}\) diverges (harmonic series).
This shows the converse is false.
But \(\sum |a_n| = \sum \frac{1}{n}\) diverges (harmonic series).
This shows the converse is false.
Key Concepts
Absolute ConvergenceComparison Testp-SeriesDivergent Series
Absolute Convergence
In mathematical terms, absolute convergence happens when the series \( \sum |a_n| \) converges. This means we're looking at the absolute value of each term in the series, and if the series of these absolute values sums to a finite number, the original series \( \sum a_n \) is absolutely convergent.
Convergence is stronger when it's absolute. If a series is absolutely convergent, then it's guaranteed to be convergent in the normal sense, but not the other way around. This is significant because it allows more powerful mathematical tools to be used, such as the rearrangement of series without affecting their sum.
For example, in the exercise, it demonstrates that if \( \sum a_n \) converges absolutely, then \( \sum a_n^2 \) will also converge because it's part of the property of absolute convergence. This feature is particularly useful in various fields of math, especially when dealing with functions and real-world data.
Convergence is stronger when it's absolute. If a series is absolutely convergent, then it's guaranteed to be convergent in the normal sense, but not the other way around. This is significant because it allows more powerful mathematical tools to be used, such as the rearrangement of series without affecting their sum.
For example, in the exercise, it demonstrates that if \( \sum a_n \) converges absolutely, then \( \sum a_n^2 \) will also converge because it's part of the property of absolute convergence. This feature is particularly useful in various fields of math, especially when dealing with functions and real-world data.
Comparison Test
The comparison test is a handy tool in the world of series to determine convergence. It involves comparing a given series with another series whose convergence behavior is known. If \( 0 \leq b_n \leq c_n \) for every term \( n \), and it is known that \( \sum c_n \) converges, then \( \sum b_n \) must also converge.
In our exercise's context, this is used to prove part (a). Since \( a_n^2 \leq |a_n|^2 \), and because \( \sum |a_n| \) converges, \( \sum |a_n|^2 \) converges automatically by the properties of series. Applying the comparison test here ensures that \( \sum a_n^2 \) converges.
The comparison test works particularly well with series that involve non-negative terms, making it ideal for dealing with absolute values and other positive sequences.
In our exercise's context, this is used to prove part (a). Since \( a_n^2 \leq |a_n|^2 \), and because \( \sum |a_n| \) converges, \( \sum |a_n|^2 \) converges automatically by the properties of series. Applying the comparison test here ensures that \( \sum a_n^2 \) converges.
The comparison test works particularly well with series that involve non-negative terms, making it ideal for dealing with absolute values and other positive sequences.
p-Series
A p-series is a series of the form \( \sum \frac{1}{n^p} \). These series converge or diverge based on the value of \( p \). Specifically:
In the exercise, the series \( \sum \frac{1}{\sqrt{n}} \) is a p-series with \( p = \frac{1}{2} \). Since \( p \) is less than 1, this specific p-series diverges. The rule for p-series comes in handy when recognizing patterns in series like these and applying this knowledge to understand their convergence or divergence.
Knowing about p-series helps in predicting the behavior of similar series, which is useful across many mathematical disciplines.
- If \( p > 1 \), the series converges.
- If \( 0 < p \leq 1 \), the series diverges.
In the exercise, the series \( \sum \frac{1}{\sqrt{n}} \) is a p-series with \( p = \frac{1}{2} \). Since \( p \) is less than 1, this specific p-series diverges. The rule for p-series comes in handy when recognizing patterns in series like these and applying this knowledge to understand their convergence or divergence.
Knowing about p-series helps in predicting the behavior of similar series, which is useful across many mathematical disciplines.
Divergent Series
A series diverges if its terms do not approach a finite sum as more terms are added. Unlike convergent series, where partial sums become increasingly close to a finite limit, divergent series don't settle down to a single value.
The exercise illustrates this with the example of \( \sum \frac{1}{n} \), which is a classic case of a divergent harmonic series. Despite the decreasing nature of its terms, the overall series grows indefinitely.
Another example is the series of absolute values \( \sum |a_n| \) from the exercise, where \( a_n = (-1)^n \frac{1}{\sqrt{n}} \). This is shown to diverge as it corresponds to a divergent p-series with \( p = \frac{1}{2} \).
Understanding divergence helps in identifying when a series will fail to converge, providing critical insights into its behavior, whether in pure math or applied contexts.
The exercise illustrates this with the example of \( \sum \frac{1}{n} \), which is a classic case of a divergent harmonic series. Despite the decreasing nature of its terms, the overall series grows indefinitely.
Another example is the series of absolute values \( \sum |a_n| \) from the exercise, where \( a_n = (-1)^n \frac{1}{\sqrt{n}} \). This is shown to diverge as it corresponds to a divergent p-series with \( p = \frac{1}{2} \).
Understanding divergence helps in identifying when a series will fail to converge, providing critical insights into its behavior, whether in pure math or applied contexts.
Other exercises in this chapter
Problem 37
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Let \(f(x)=\sum_{n=1}^{\infty} \frac{(x-2)^{n}}{n^{2} 3^{n}}\). Show that the domain of \(f\) is \([-1,5]\) but the domain of \(f^{\prime}\) is \([-1,5)\).
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