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Chapter 3: Problem 8

If \(\sum a_{n}\) with \(a_{n}>0\) is convergent, then is \(\sum a_{n}^{2}\) always convergent? Either prove it or give a counterexample.

Short Answer

Expert verified
No, \(\sum a_{n}^{2}\) is not always convergent. For example, when \(a_{n}=1/n\), \(\sum a_{n}\) is convergent, but \(\sum a_{n}^{2}\) is divergent.

Step by step solution

01

Understanding the Problem

The exercise wants us to determine whether the series \(\sum a_{n}^{2}\) is convergent given the fact that \(\sum a_{n}\) is convergent and each term \(a_{n}\) is greater than 0.
02

Providing a Counterexample

Upon analyzing, it can be determined that the statement can be false. Consider the sequence where \(a_{n} = 1/n\). It is a positive sequence and \(\sum a_{n}\) converges (it is basically the harmonic series). However, \(\sum a_{n}^{2} = \sum 1/n^{2}\) diverges, because the sum of the reciprocal of any series that converges is usually divergent. Hence, this is a counterexample of the statement.

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Most popular questions from this chapter

Show that if \(\left(x_{n}\right)\) is unbounded, then there exists a subsequence \(\left(x_{n_{k}}\right)\) such that \(\lim \left(1 / x_{n_{2}}\right)=0\).

If \(x_{1}>0\) and \(x_{n+1}:=\left(2+x_{n}\right)^{-1}\) for \(n \geq 1\), show that \(\left(x_{n}\right)\) is a contractive sequence. Find the limit.

Use the Cauchy Condensation Test to discuss the \(p\) -series \(\sum_{n=1}^{\infty}\left(1 / n^{p}\right)\) for \(p>0\).

Can you give an example of a convergent series \(\sum x_{n}\) and a divergent series \(\sum y_{n}\) such that \(\sum\left(x_{n}+y_{n}\right)\) is convergent? Explain.

Let \(\left(x_{n}\right)\) and \(\left(y_{n}\right)\) be sequences of positive numbers such that \(\lim \left(x_{n} / y_{n}\right)=0\). (a) Show that if \(\lim \left(x_{n}\right)=+\infty\), then \(\lim \left(y_{n}\right)=+\infty\). (b) Show that if \(\left(y_{n}\right)\) is bounded, then \(\lim \left(x_{n}\right)=0\).
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