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Chapter 3: Problem 5

If \(x_{n}:=\sqrt{n}\), show that \(\left(x_{n}\right)\) satisfies \(\lim \left|x_{n+1}-x_{n}\right|=0\), but that it is not a Cauchy sequence.

Short Answer

Expert verified
The sequence \(x_{n}:=\sqrt{n}\) satisfies \(\lim_{n \to \infty} \left|x_{n+1} - x_{n}\right| = 0\), but it is not a Cauchy sequence since the differences \(|x_{n+1} - x_n|\) can not be made as small as we like, however far out in the sequence we go.

Step by step solution

01

Understanding the Provided Sequence

Firstly, analyze the provided sequence. The sequence is \(x_{n}:=\sqrt{n}\). This means each term in the sequence is the square root of its place in the sequence, i.e. \(x_{1} = \sqrt{1}\), \(x_{2} = \sqrt{2}\), \(x_{3} = \sqrt{3}\), etc.
02

Calculating the Limit of the Absolute Difference

By the problem's statement, \(lim_{n \to \infty} \left|x_{n+1} - x_{n}\right| = 0\). This means the absolute difference between consecutive terms of the sequence decreases as to approach zero as we go further along the sequence. To prove this, let's compute this limit using the formula: \[lim_{n \to \infty} \left|\sqrt{n+1} - \sqrt{n}\right| = lim_{n \to \infty} \frac {|\sqrt{n+1} - \sqrt{n}| ^ 2} {|\sqrt{n+1} + \sqrt{n}|}= lim_{n \to \infty} \frac {n+1 - n} {|\sqrt{n+1} + \sqrt{n}|}=lim_{n \to \infty} \frac {1} {|\sqrt{n+1} + \sqrt{n}|}=0\]so indeed, \(\left|x_{n+1} - x_{n}\right|\) has a limit of 0.
03

Identifying Criteria for a Cauchy Sequence

A sequence is a Cauchy sequence if, and only if, for any small positive number \( \epsilon \), there exists a positive integer \( N \) such that for all \( n, m \) greater than \( N \), the absolute difference between the \( n^{th} \) and \( m^{th} \) terms of the sequence is less than \( \epsilon \). More formally, if \( \forall \epsilon > 0 \, \exists N 鈭圽mathbb{Z}^+ \, \forall m,n > N: \left| x_m - x_n \right| < \epsilon \), where \( x_n \) is the \( n^{th} \) term of the sequence.
04

Checking if Sequence is a Cauchy Sequence

Let鈥檚 test if the sequence \(x_{n}:=\sqrt{n}\) is a Cauchy sequence. If we take \( \epsilon = 1/4 \), and try to find an \( N \) such that \( \forall m,n > N, \left| x_m - x_n \right| < \epsilon \), we see that it鈥檚 not possible since:\[|x_{n+1} - x_n| = |\sqrt{n+1} - \sqrt{n}| > 1/4 \quad \forall n \geq 1\]So, however large \( N \) might be, there would always exist \( m = N + 1 \) and \( n = N \) such that the absolute difference is not less than \( \epsilon \). Thus, the sequence \(x_{n}\) is not a Cauchy sequence.

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Most popular questions from this chapter

Show that \(\lim \left(2^{n} / n !\right)=0 .\) [Hint: if \(n \geq 3\), then \(0<2^{n} / n ! \leq 2\left(\frac{2}{3}\right)^{n-2}\).]

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

The sequence \(\left(x_{n}\right)\) is defined by che following formulas for the \(n\) th term. Write the furst five terms in each case: (a) \(x_{n}:=1+(-1)^{n}\), (b) \(x_{n}:=(-1)^{n} / n\), (c) \(x_{n}:=\frac{1}{n(n+1)}\), (d) \(x:=\frac{1}{n^{2}+2}\).

By using partial fractions, show that (a) \(\sum_{n=0}^{\infty} \frac{1}{(n+1)(n+2)}=1\), (b) \(\sum_{n=0}^{\infty} \frac{1}{(\alpha+n)(\alpha+n+1)}=\frac{1}{\alpha}>0\), if \(\alpha>0\). (c) \(\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)}=\frac{1}{4}\).
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