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Chapter 9: Problem 5

If \(\sum a_{n}\) is absolutely convergent, is it true that every rearrangement of \(\sum a_{n}\) is also absolutely convergent?

Short Answer

Expert verified
Yes, every rearrangement of an absolutely convergent series \(\sum a_{n}\) will also be absolutely convergent. This is result of the Riemann Rearrangement Theorem which states that a series is absolutely convergent if and only if every rearrangement of it is also absolutely convergent.

Step by step solution

01

Define the Problem

Check if every rearrangement of an absolutely convergent series \(\sum a_{n}\) is also absolutely convergent.
02

Statement of Theorem

For any absolutely convergent series, rearrangements will also retain absolute convergence. This is called the Riemann Rearrangement Theorem.
03

Rationale

Let \(\sum a_{n}\) be an absolutely convergent series. Then, \(\sum |a_{n}| < 鈭瀄). Even if the terms are rearranged in a different order, the sum of their absolute values would still be less than infinity, as the sum is independent of the order of terms. Therefore, a rearrangement of the series is also absolutely convergent.

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Most popular questions from this chapter

Show that if \(|x|<1\), then \(\operatorname{Arcsin} x=\sum_{n=0}^{\infty} \frac{1 \cdot 3 \cdots(2 n-1)}{2 \cdot 4 \cdots 2 n} \cdot \frac{x^{2 n+1}}{2 n+1}\).

Discuss the convergence or the divergence of the series with \(n\) th term (for sufficiently Iarge \(n\) ) given by (a) \((\ln n)^{-p}\), (b) \((\ln n)^{-n}\) (d) \((\ln n)^{-\ln \ln n}\) (c) \((\ln n)^{-\ln n}\), (e) \((n \ln n)^{-1}\), (f) \(\left(n(\ln n)(\ln \ln n)^{2}\right)^{-1}\).

Prove that if \(f\) is defined for \(|x|

Show that the series \(1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}++-\cdots\) is divergent.

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