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Chapter 3: Problem 4

Show that the following sequences are divergent. (a) \(\left(1-(-1)^{n}+1 / n\right)\), (b) \((\sin n \pi / 4)\).

Short Answer

Expert verified
Both sequences are divergent: (a) \(\left(1-(-1)^{n}+1 / n\right)\) due to its alternating nature between two values, and (b) \((\sin n \pi / 4)\) because it oscillates between several values.

Step by step solution

01

Analyze Sequence (a)

Sequnce (a), \(\left(1-(-1)^{n}+1 / n\right)\), oscillates due to the term \((-1)^{n}\). This part of the sequence goes to -1 when n is odd and 1 when n is even.
02

Determine Limit of Sequence (a)

Since the sequence alternates between two values when \(n\) is an odd or an even number, it does not have a obtainable limit, implying that the sequence is divergent.
03

Analyze Sequence (b)

Sequence (b), \((\sin n \pi / 4)\), fluctuates in a predictable pattern over time. When \(n\) is increased by 1, the argument of the sine function is increased by \(\pi / 4\). This results in eight distinct values, after which the sequence repeats itself.
04

Determine Limit of Sequence (b)

The sequence does not approach a single real number as \(n\) tends to infinity. Instead, it oscillates between several values. This reveals that sequence (b) is also divergent.

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Most popular questions from this chapter

Show that the following sequences are not convergent. (a) \(\left(2^{n}\right)\), (b) \(\left((-1)^{n} n^{2}\right)\).

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

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

Show that if \(\left(x_{n}\right)\) is an unbounded sequence, then there exists a properly divergent subsequence.

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) \(s_{1}=3, \quad s_{2}:=5, \quad s_{n+2}:=s_{n}+s_{n+1}\).
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