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Chapter 3: Problem 5

Is the sequence \((n \sin n)\) properly divergent?

Short Answer

Expert verified
Yes, the sequence \(n \sin n\) is properly divergent as it oscillates without bound in both the positive and negative directions.

Step by step solution

01

Understanding the sequence elements

First, let's analyze the elements in this sequence. For each term \(a_n\) of the sequence, it is the product of \(n\) (which increases with each term) and \(\sin{n}\), a sine function that continuously oscillates between -1 and 1. As we increase \(n\), the product \(n \sin n\) thus can become large positive, large negative, or near zero, depending on the value of \(\sin n\).
02

Identifying an oscillating sequence.

Since \(\sin n\) oscillates between -1 and 1 for all real numbers n, there are infinitely many values of n where \(\sin n = 1\) and \(\sin n = -1\). Therefore, \(n \sin n\) will take on values as high as \(n\) (when \(\sin n = 1\)) and as low as \(-n\) (when \(\sin n = -1\)). Because \(n\) goes to infinity, so do these maximum and minimum values, both in the positive and negative directions.
03

Concluding the nature of divergence

The sequence \(n \sin n\) therefore does not converge to a finite limit or diverge to positive or negative infinity. Rather, it oscillates without bound in both the positive and negative directions, meaning the sequence is properly divergent.

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Most popular questions from this chapter

Suppose that \(\left(x_{n}\right)\) is a convergent sequence and \(\left(y_{n}\right)\) is such that for any \(\varepsilon>0\) there exists \(M\) such that \(\left|x_{n}-y_{n}\right|<\varepsilon\) for all \(n \geq M\). Does it follow that \(\left(y_{n}\right)\) is convergent?

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

Let \(\left(x_{n}\right)\) be a Cauchy sequence such that \(x_{n}\) is an integer for every \(n \in \mathbb{N}\). Show that \(\left(x_{n}\right)\) is. ultimately constant.

Show that if \(X\) and \(Y\) are sequences such that \(X\) and \(X+Y\) are convergent, then \(Y\) is convergent.

Show that \(\lim \left(n^{2} / n !\right)=0\).
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