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The question has two parts

Let $x_n$ be a sequence where $x_n=\sqrt{n}$

a) Is $\{x_n\}$ a contractive sequence?

b) Prove that $\frac{|x_{n+2}-x_{n+1}|}{|x_{n+1}-x_n|}=\frac{|\sqrt{n+2}-\sqrt{n+1}|}{|\sqrt{n+1}-\sqrt{n}|}<1$ for all n.

Notes:

A) $\{x_n\}$ has no upper bound since for any possible integer $q$, there is a greater integer (e.g. $\sqrt{(q+1)^2}>\sqrt{q}$). If $\{x_n\}$ was contractive, it would also be Cauchy and converge (as $\mathbb{R}$ is complete). Therefore it would be bounded, thus we have a contradiction so $\{x_n\}$ is not contractive.

B) Simply graphing $\sqrt{n+1}-\sqrt{n}$ shows that it is a decreasing sequence. But I am not sure how to prove so using algebra.

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AzJ
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1 Answers1

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Using the Square Root Inequality,

$$\sqrt{n+2}-\sqrt{n+1} < \frac{1}{2\sqrt{n+1}} < \sqrt{n+1} - \sqrt{n}$$

Therefore,

$${\sqrt{n+2}-\sqrt{n+1} \over \sqrt{n+1}-\sqrt{n}} < 1$$

See the following for a proof of the inequality:

Proof of inequality $2(\sqrt{n+1}-\sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n-1})$ using induction

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Alexis Olson
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  • Thank you for the answering part B. However how does the statement ${\sqrt{n+2}-\sqrt{n+1} \over \sqrt{n+1}-\sqrt{n}} < 1$ show that $x_n$ is contractive. I feel like there is something I am missing. – AzJ Oct 04 '16 at 17:54
  • @NhG I'm sorry, you are correct. It is not contractive. – Alexis Olson Oct 04 '16 at 17:57