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Chapter 3: Problem 4

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

Short Answer

Expert verified
All four sequences properly diverge to infinity. The limit of \((\sqrt{n})\), \((\sqrt{n+1})\), and \((\sqrt{n-1})\) as \(n\) approaches infinity is infinity. Similarly, the limit of \((n / \sqrt{n+1})\) as \(n\) approaches infinity is also infinity.

Step by step solution

01

Analysing sequence (a) \((\sqrt{n})\)

We need to calculate the limit as \(n\) approaches infinity for \((\sqrt{n})\). By direct inspection, as \(n\) gets larger and larger, \(\sqrt{n}\) also increases. So, we can say that \( \lim_{n \to \infty} \sqrt{n} = \infty \).
02

Analysing sequence (b) \((\sqrt{n+1})\)

Just like in step one, we are calculating the limit as \(n\) approaches infinity for \((\sqrt{n+1})\). Given that we're adding 1 to \(n\), the sequence increases even quicker than \(\sqrt{n}\). Therefore, \( \lim_{n \to \infty} \sqrt{n+1} = \infty \).
03

Analysing sequence (c) \((\sqrt{n-1})\)

We're calculating the limit as \(n\) approaches infinity for \((\sqrt{n-1})\). Even though we're subtracting 1 from \(n\), the sequence will still increase as \(n\) increases, albeit slightly slower than \(\sqrt{n}\). Still, \( \lim_{n \to \infty} \sqrt{n-1} = \infty \).
04

Analysing sequence (d) \((n / \sqrt{n+1})\)

To solve this, we calculate the limit as \(n\) approaches infinity for \((n / \sqrt{n+1})\). Here, we divide \(n\) by \(\sqrt{n+1}\), so the sequence increases at a different rate than the previous sequences. It's important to note that the denominator increases at a slower rate than the numerator as \(n\) increases. As such, \( \lim_{n \to \infty} (n / \sqrt{n+1}) = \infty \).

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Most popular questions from this chapter

Show that (a) \(\lim \left(\frac{1}{\sqrt{n+7}}\right)=0\), (b) \(\lim \left(\frac{2 n}{n+2}\right)=2\), (c) \(\lim \left(\frac{\sqrt{n}}{n+1}\right)=0\), (d) \(\lim \left(\frac{(-1)^{n} n}{n^{2}+1}\right)=0\).

Show that \(\lim \left(\frac{1}{n}-\frac{1}{n+1}\right)=0\).

Give examples of properly divergent sequences \(\left(x_{n}\right)\) and \(\left(y_{n}\right)\) with \(y_{n} \neq 0\) for all \(n \in \mathbb{N}\) such that: (a) \(\left(x_{n} / y_{n}\right)\) is convergent, (b) \(\left(x_{n} / y_{n}\right)\) is properly divergent.

Can you give an example of a convergent series \(\sum x_{n}\) and a divergent series \(\sum y_{n}\) such that \(\sum\left(x_{n}+y_{n}\right)\) is convergent? Explain.

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