Answer to Question #320263 in Real Analysis for Akashdip

Question

Using the definition, show' that the sequence [1/√n]neN is Cauchy.

Answer 1

ANSWER.

By the definition, a sequence $\left \{a _{n} \right \}$ is called a Cauchy sequence if for any given $\varepsilon>0$ there exists $n_{\varepsilon}\in N$ such that

$n,m>n_{\varepsilon }\Rightarrow\left | a_{n} -a_{m}\right |<\varepsilon$ .

Note, that if $m \neq n$ , then $m>n$ (or vice versa). Let $m>n$ , then $m=n+(m-n)$ Denoting $p=m-n$, we get the equivalent definition: $\left \{a _{n} \right \}$ is called a Cauchy

sequence if for any given $\varepsilon>0$ there exists $n_{\varepsilon}\in N$ such that

$\left | a_{n} -a_{n+p}\right |<\varepsilon$ for all $n >n_{\varepsilon }$ and $p\in N$ .

Since $\sqrt{n+p}>\sqrt{n}$ for all $n,p\in N$ ,then

$\left | a_{n}-a_{n+p} \right |=\frac{1}{\sqrt{n }}- \frac{1}{\sqrt{n+p}}<\frac{1}{\sqrt{n }}$ .

So, if $\varepsilon>0$ and $n_{\varepsilon}=$$\left [ \left ( \frac{1}{\varepsilon} \right )^{2} \right ]$ (integer part of $\left ( \frac{1}{\varepsilon} \right )^{2}$ ) , then

$n>n_{\varepsilon}>\left ( \frac{1}{\varepsilon} \right )^{2} \Rightarrow \sqrt{n}>\sqrt{n_{\varepsilon}} >\left ( \frac{1}{\varepsilon} \right ) \Rightarrow0< \frac{1}{\sqrt{n}}<\varepsilon$ ,

Hense , $0<\frac{1}{\sqrt{n }}- \frac{1}{\sqrt{n+p}}<\varepsilon$ $(n>n_{\varepsilon}, p\geq1)$ .

Therefore , the sequence $a_{n}=\frac{1}{\sqrt{n}}$ is Cauchy sequence