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Chapter 5: Problem 235

Show that if \(a_{n} \geq 0\) and \(\sum_{n=1}^{\infty} a_{n}\) converges, then \(\sum_{n=1}^{\infty} a_{n}^{2}\) converges. If \(\sum_{n=1}^{\infty} a_{n}^{2}\) converges, does \(\sum_{n=1}^{\infty} a_{n}\) necessarily converge?

Short Answer

Expert verified
Yes, if \(\sum a_n\) converges, \(\sum a_n^2\) converges; converse is false.

Step by step solution

01

Analyze the conditions and premises

We start by considering the given information that the sequence \(a_n\) is non-negative, \(a_n \geq 0\), and the series \(\sum_{n=1}^{\infty} a_n\) converges. We need to show that \(\sum_{n=1}^{\infty} a_n^2\) also converges under these conditions.
02

Apply properties of convergent series

We know that if a series \(\sum_{n=1}^{\infty} a_n\) converges, then the terms \(a_n\) must converge to zero, \(a_n \to 0\). This implies that for sufficiently large \(n\), \(a_n\) becomes arbitrarily small.
03

Comparison test setup

Consider \(a_n^2\). Because \(0 \leq a_n < 1\) for sufficiently large \(n\) due to convergence, we have \(a_n^2 < a_n\). Since \(\sum_{n=1}^{\infty} a_n\) converges, by the Comparison Test, \(\sum_{n=1}^{\infty} a_n^2\) must also converge.
04

Counterexample for the converse

To determine if the converse is true, i.e., if \(\sum_{n=1}^{\infty} a_n^2\) converges implies \(\sum_{n=1}^{\infty} a_n\) also converges, consider \(a_n = \frac{1}{\sqrt{n}}\). Here, \(a_n^2 = \frac{1}{n}\), which converges (known harmonic series variant), yet \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}\) diverges. Thus, the converse does not hold.

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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 valuable tool in determining the convergence of a series by comparing it to another series whose convergence is already known. This test is particularly useful when dealing with series that are challenging to assess directly. When attempting to use the Comparison Test, follow these guidelines:
  • Identify the series of interest, which we'll call \sum_{n=1}^{\infty} b_n\, and compare it to a known convergent or divergent series, \sum_{n=1}^{\infty} a_n\.
  • Ensure that the terms satisfy \(0 \leq b_n \leq a_n\) for all \(n\) from some initial index onward.
  • If \sum_{n=1}^{\infty} a_n\ converges and \sum_{n=1}^{\infty} b_n\ is bounded by it as stated, the series \sum_{n=1}^{\infty} b_n\ must also converge.
Applying this method in the context of the original exercise, the given series \sum_{n=1}^{\infty} a_n^2\ was compared to \sum_{n=1}^{\infty} a_n\. Since \(a_n^2 \leq a_n\), and the latter converges, the Comparison Test confirms the convergence of the former.
Convergent Series
A series \sum_{n=1}^{\infty} a_n\ is said to converge when its sequence of partial sums approaches a finite limit as \(n\) tends to infinity. Essentially, this means that adding more and more terms from \(a_n\) results in a sum that stabilizes and stops growing indefinitely.Several key points characterize convergent series:
  • The terms \(a_n\) must approach zero as \(n\) becomes large; otherwise, they will not converge.
  • The partial sums \(s_n = a_1 + a_2 + ... + a_n\) eventually settle to a certain value.
  • Convergence is often easier to determine using tests such as the Comparison Test, Ratio Test, or Integral Test, among others.
The exercise highlights that if \sum_{n=1}^{\infty} a_n\ converges, then \sum_{n=1}^{\infty} a_n^2\ will also converge due to the relation \(a_n^2 < a_n\) for sufficiently large \(n\).
Counterexample
Counterexamples are a powerful way to demonstrate that a particular statement does not universally hold true. By presenting a specific case where the conditions are met, but the expected outcome does not occur, we can effectively invalidate a generalized assertion.In the original exercise, the statement that the convergence of \sum_{n=1}^{\infty} a_n^2\ implies the convergence of \sum_{n=1}^{\infty} a_n\ is challenged. Here is a notable counterexample:
  • Consider \(a_n = \frac{1}{\sqrt{n}}\).
  • This gives \(a_n^2 = \frac{1}{n}\), a series known to converge.
  • However, \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}\, which is the series for \sum_{n=1}^{\infty} a_n\, diverges.
Thus, the original claim cannot be assumed true universally, highlighting the importance of counterexamples in mathematical analysis.
Series Analysis
Series analysis involves an in-depth examination of the behavior and properties of infinite series. This type of analysis is crucial in fields such as mathematics and physics, where understanding infinite sums is fundamental.Several aspects make up effective series analysis:
  • Convergence/Divergence: Checking whether a series converges or diverges, using various tests like the Comparison Test, often unveils deeper insights.
  • Properties of Terms: Understanding how the terms of a series behave, especially as \(n\) becomes large, aids in assessing the series as a whole.
  • Pattern Recognition: Often, recognizing a familiar pattern or known series within a new problem simplifies the analysis.
Incorporating these approaches, the exercise walks through using the Comparison Test and examining the behavior of series \(a_n^2\) to demonstrate its convergence when \(a_n\) does, while also using a counterexample to refute the reverse implication. This embodies the holistic study needed in series analysis.

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Most popular questions from this chapter

Let \(a_{n}=2^{-[n / 2]}\) where \([x]\) is the greatest integer less than or equal to \(x\). Determine whether \(\sum_{n=1}^{\infty} a_{n}\) converges and justify your answer. The following advanced exercises use a generalized ratio test to determine convergence of some series that arise in particular applications when tests in this chapter, including the ratio and root test, are not powerful enough to determine their convergence. The test states that if \(\lim _{n \rightarrow \infty} \frac{a_{2 n}}{a_{n}}<1 / 2\), then \(\sum a_{n}\) converges, while if \(\lim _{n \rightarrow \infty} \frac{a_{2 n+1}}{a_{n}}>1 / 2\), then \(\sum a_{n}\) diverges.

[T] The Euler transform rewrites \(S=\sum_{n=0}^{\infty}(-1)^{n} b_{n}\) as \(S=\sum_{n=0}^{\infty}(-1)^{n} 2^{-n-1} \sum_{m=0}^{n}\left(\begin{array}{l}n \\\ m\end{array}\right) b_{n-m} . \quad\) For the alternating harmonic series, it takes the form \(\ln (2)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}=\sum_{n=1}^{\infty} \frac{1}{n 2^{n}} . \quad\) Compute \(\quad\) partial sums of \(\sum_{n=1}^{\infty} \frac{1}{n 2^{n}}\) until they approximate \(\ln (2)\) accurate to within 0.0001 . How many terms are needed? Compare this answer to the number of terms of the alternating harmonic series are needed to estimate \(\ln (2)\).

State whether each of the following series converges absolutely, conditionally, or not at all. $$\sum_{n=1}^{\infty} \frac{\cos (n \pi)}{n^{1 / n}}$$

Use the ratio test to determine whether \(\sum_{n=1}^{\infty} a_{n}\) converges, where \(a_{n}\) is given in the following problems. State if the ratio test is inconclusive. $$\sum_{n=1}^{\infty} \frac{(2 n) !}{(n / e)^{2 n}}$$

State whether each of the following series converges absolutely, conditionally, or not at all. $$\sum_{n=1}^{\infty}(-1)^{n+1}\left((n+1)^{2}-n^{2}\right)$$
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