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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 the convergence of a series is not affected by changing a finite number of its terms. (Of course, the value of the sum may be changed.)

Use the Cauchy Condensation Test to discuss the \(p\) -series \(\sum_{n=1}^{\infty}\left(1 / n^{p}\right)\) for \(p>0\).

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.

Show directly from the definition that the following are Cauchy sequences. (a) \(\left(\frac{n+1}{n}\right)\). (b) \(\left(1+\frac{1}{2 !}+\cdots+\frac{1}{n !}\right)\).

. Let \(\left(x_{n}\right)\) be a bounded sequence and let \(s:=\sup \left(x_{n}: n \in \mathbb{N}\right)\). Show that if \(s \notin\left\\{x_{n}: n \in \mathbb{N}\right\\}\), then there is a subsequence of \(\left(x_{n}\right)\) that converges to \(s\).
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