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Chapter 3: Problem 6

(a) Show that the series \(\sum_{n=1}^{\infty} \cos n\) is divergent. (b) Show that the series \(\sum_{n=1}^{\infty}(\cos n) / n^{2}\) is convergent.

Short Answer

Expert verified
The series \(\sum_{n=1}^\infty \cos n\) is divergent while the series \(\sum_{n=1}^\infty (\cos n) / n^{2}\) is convergent.

Step by step solution

01

Understanding Divergent and Convergent Series

A series \(\sum_{n=1}^\infty a_n\) is said to converge if the sequence of its partial sums \(S_n = \sum_{i=1}^n a_i\) approaches a limit as n approaches infinity; otherwise, the series is said to diverge.
02

Showing Divergence for \(\sum_{n=1}^\infty \cos n\)

We know that the sum of the series is the limit of the sequence of its partial sums as n goes to infinity, i.e., \(\lim_{n\to\infty} S_n = \lim_{n\to\infty} \sum_{i=1}^n \cos(i)\). Here, the individual terms do not approach zero as n approaches infinity, thus we can't establish the series converges. Therefore, this series is divergent.
03

Showing Convergence for \(\sum_{n=1}^\infty (\cos n) / n^{2}\)

To ascertain this, we'll implement the comparison test. We know that for positive series, if 0 <= \(a_n\) <= \(b_n\) for all n, then if \(b_n\) converges, so does \(a_n\), and if \(a_n\) diverges, so does \(b_n\). In our case, we are comparing with \(\sum_{n=1}^\infty 1/n^2\), which is a known convergent series (p-series with p=2 > 1). Since the absolute value of each term in \(\sum_{n=1}^\infty (\cos n) / n^{2}\) is less than or equal to each corresponding term in \(\sum_{n=1}^\infty 1/n^2\), the series \(\sum_{n=1}^\infty (\cos n) / n^{2}\) converges.

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Most popular questions from this chapter

Establish the proper divergence of the following sequences. (a) \((\sqrt{n})\), (b) \((\sqrt{n+1})\), (c) \((\sqrt{n-1})\) (d) \((n / \sqrt{n+1})\).

Let \(y_{n}:=\sqrt{n+1}-\sqrt{n}\) for \(n \in \mathbb{N}\). Show that \(\left(y_{n}\right)\) and \(\left(\sqrt{n} y_{n}\right)\) converge. Find their limits.

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 .

Let \(\left(x_{n}\right)\) be a bounded sequence and for each \(n \in \mathbb{N}\) let \(s_{n}:=\sup \left\\{x_{k}: k \geq n\right\\}\) and \(S:=\inf \left\\{s_{n}\right\\}\). Show that there exists a subsequence of \(\left(x_{n}\right)\) that converges to \(S .\)

Show directly from the definition that if \(\left(x_{n}\right)\) and \(\left(y_{n}\right)\) are Cauchy sequences, then \(\left(x_{n}+y_{n}\right)\) and \(\left(x_{n} y_{n}\right)\) are Cauchy sequences.
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