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Chapter 3: Problem 10

If \(\sum a_{n}\) with \(a_{n}>0\) is convergent, then is \(\sum \sqrt{a_{n} a_{n+1}}\) always convergent? Either prove it or give a counterexample.

Short Answer

Expert verified
No, it's not always the case that if \( \sum a_{n} \) with \( a_{n} > 0 \) is convergent, then \( \sum \sqrt{a_{n} a_{n+1}} \) is also convergent. The counterexample provided, where \( a_{n} = \frac{1}{n^{2}} \), indicates that there are sequences that violate this condition.

Step by step solution

01

Understand the series

The statement is saying that if \( \sum a_{n} \) with \( a_{n} > 0 \) is convergent, then \( \sum \sqrt{a_{n} a_{n+1}} \) is always convergent. A convergent series is a series that has a finite sum.
02

Consider a counterexample

Consider a sequence \( a_{n} = \frac{1}{n^{2}} \). This sequence is obviously positive and its series \( \sum \frac{1}{n^{2}} \) is convergent. However, the series \( \sum \sqrt{a_{n} a_{n+1}} = \sum \sqrt{ \frac{1}{n^2} \cdot \frac{1}{(n+1)^2} } \) simplifies to \( \sum \frac{1}{n(n+1)} \).
03

Analyze the counterexample

Now, splitting up the terms in the sum yields \( \sum \left(\frac{1}{n} - \frac{1}{n+1} \right) \). This series is telescopic, which means that a lot of the terms cancel out when we expand the series. However, in this case, the series diverges because the terms \( \frac{1}{n} \) do not approach zero as \( n \) tends to infinity.

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Most popular questions from this chapter

Give an example of a bounded sequence that is not a Cauchy sequence. i \(\cdots\)

If \(\sum a_{n}\) with \(a_{n}>0\) is convergent, and if \(b_{n}:=\left(a_{1}+\cdots+a_{n}\right) / n\) for \(n \in \mathbb{N}\), then show that \(\sum b_{n}\) is always divergent.

Let \(\left(x_{n}\right)\) be properly divergent and let \(\left(y_{n}\right)\) be such that \(\lim \left(x_{n} y_{n}\right)\) belongs to \(\mathbb{R}\). Show that \(\left(y_{n}\right)\) converges to 0 .

List the first five terms of the following inductively defined sequences. (a) \(x_{1}:=1, \quad x_{n+1}=3 x_{n}+1\), (b) \(y_{1}:=2, \quad y_{n+1}=\frac{1}{2}\left(y_{n}+2 / y_{n}\right)\) (c) \(z_{1}:=1, \quad z_{2}:=2, \quad z_{n+2}:=\left(z_{n+1}+z_{n}\right) /\left(z_{n+1}-z_{n}\right)\) (d) \(s_{1}=3, \quad s_{2}:=5, \quad s_{n+2}:=s_{n}+s_{n+1}\).

If \(x_{1}2\), show that \(\left(x_{n}\right)\) is convergent. What is its limit?
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