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Chapter 9: Problem 7

(a) If \(\sum a_{n}\) is absolutely convergent and \(\left(b_{n}\right)\) is a bounded sequence, show that \(\sum a_{n} b_{n}\) is absolutely convergent. (b) Give an example to show that if the convergence of \(\sum a_{n}\) is conditional and \(\left(b_{n}\right)\) is a bounded sequence, then \(\sum a_{n} b_{n}\) may diverge.

Short Answer

Expert verified
For part (a), the series \(\sum a_{n} b_{n}\) is absolutely convergent given that \(\sum a_{n}\) is absolutely convergent and \((b_{n})\) is a bounded sequence. For part (b), an example of a divergent series even though \(\sum a_{n}\) is conditionally convergent and \(b_{n}\) is bounded, can be illustrated by \(a_{n} = (-1)^{n} / n\) and \(b_{n} = (-1)^{n}\).

Step by step solution

01

Prove the absolute convergence

Since \(\sum a_{n}\) is absolutely convergent, \(\sum |a_{n}|\) is convergent. Let M be an upper bound of the sequence \(b_{n}\). Now consider the series \(\sum |a_{n} b_{n}|\). Since \(|b_{n}|\) is less than or equal to M, it follows that \(|a_{n} b_{n}|\) is less than or equal to \(M |a_{n}|\). Therefore, using comparison test, we have \(\sum |a_{n} b_{n}|\) is convergent. Hence, \(\sum a_{n} b_{n}\) is absolutely convergent.
02

Construct an example showing divergence

Let \(a_{n} = (-1)^{n} / n\) and \(b_{n} = (-1)^{n}\). Then \(\sum a_{n}\) is conditionally convergent (it is the alternating harmonic series) and \(b_{n}\) is bounded by 1. Now consider the series \(\sum a_{n} b_{n} = \sum (-1)^{2n} / n = \sum 1 / n\), which is the harmonic series and is known to be divergent. Thus, \(\sum a_{n} b_{n}\) diverges even though \(\sum a_{n}\) is conditionally convergent and \(b_{n}\) is bounded.

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Most popular questions from this chapter

Let \(a: \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{R}\) and write \(a_{i j}:=a(i, j)\). If \(A_{i}:=\sum_{j=1}^{\infty} a_{i j}\) for each \(i \in \mathbb{N}\) and if \(A:=\) \(\sum_{i=1}^{\infty} A_{i}\), we say that \(A\) is an iterated sum of the \(a_{i j}\) and write \(A=\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} a_{i j} .\) We define the other iterated sum, denoted by \(\sum_{j=1}^{\infty} \sum_{i=1}^{\infty} a_{i j}\), in a similar way. Suppose \(a_{i j} \geq 0\) for \(i, j \in \mathbb{N} .\) If \(\left(c_{k}\right)\) is any enumeration of \(\left(a_{i j}: i, j \in \mathbb{N}\right\\}\), show that the following statements are equivalent: (i) The interated sum \(\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} a_{i j}\) converges to \(B\). (ii) The series \(\sum_{k=1}^{\infty} c_{k}\) converges to \(C\). In this case, we have \(B=C\).

For \(n \in \mathbb{N}\), let \(c_{n}\) be defined by \(c_{n}:=\frac{1}{1}+\frac{1}{2}+\cdots+1 / n-\ln n .\) Show that \(\left(c_{n}\right)\) is a decreasing sequence of positive numbers. The limit \(C\) of this sequence is called Euler's Constant and is approximately equal to \(0.577 .\) Show that if we put $$b_{n}:=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\cdots-\frac{1}{2 n}$$ then the sequence \(\left(b_{n}\right)\) converges to \(\ln 2 .\) [Hint: \(\left.b_{n}=c_{2 n}-c_{n}+\ln 2 .\right\\}\)

(Geometric series) Show directly that if \(|x|<1\), then \(1 /(1-x)=\sum_{n=0}^{\infty} x^{n}\).

If \(\left(a_{n}\right)\) is a bounded decreasing sequence and \(\left(b_{n}\right)\) is a bounded increasing sequence and if \(x_{n}:=a_{n}+b_{n}\) for \(n \in \mathbb{N}\), show that \(\sum_{n=1}^{\infty}\left|x_{n}-x_{n+1}\right|\) is convergent.

Let \(a_{n}>0\) and suppose that \(\sum a_{n}\) converges. Construct a convergent series \(\sum b_{n}\) with \(b_{n}>0\) such that \(\lim \left(a_{n} / b_{n}\right)=0 ;\) hence \(\sum b_{n}\) converges less rapidly than \(\sum a_{n} .\left[\right.\) Hint \(:\) Let \(\left(A_{n}\right)\) be the partial sums of \(\sum a_{n}\) and \(A\) its limit. Define \(b_{1}:=\sqrt{A}-\sqrt{A-A_{1}}\) and \(b_{n}:=\sqrt{A-A_{n-1}}-\) \(\sqrt{A-A_{n}}\) for \(\left.n \geq 1 .\right]\)
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