Show that $a_n=\frac{1}{\sqrt n}$ is cauchy directly from definition.

Question

Show that $a_n=\frac{1}{\sqrt n}$ is cauchy directly from definition.

My try:I can prove that $a_n$ converges to $0$ but how can we prove directly from definition that $a_n$ is cauchy?Thank you.

Answer 1

What you want to prove is that given $\epsilon >0$ there exist $N$ such that for any $m,n>N$ we have,

$$|a_m-a_n|< \epsilon$$

Sketch:

$$|\frac{1}{\sqrt{m}}-\frac{1}{\sqrt{n}}|< \epsilon$$

Without loss of generality, let $m \geq n$ so that by the triangle inequality,

$$|\frac{1}{\sqrt{n}}|+|\frac{1}{\sqrt{n}}| \geq |\frac{1}{\sqrt{m}}-\frac{1}{\sqrt{n}}|$$

So showing that,

$$\frac{2}{\sqrt{n}}<\epsilon$$

Is enough.

So we have,

$$\frac{2}{\epsilon}<\sqrt{n}$$

$$\frac{4}{\epsilon^2}

Proof:

Take $N=\frac{4}{\epsilon^2}$ so that $n>\frac{4}{\epsilon^2}$. Also without loss of generality take $m \geq n$ then we have,

$$|a_n-a_m| \leq |a_n|+|a_m|$$

$$\leq \frac{2}{\sqrt{n}}<\epsilon$$

Answer 2

It is easy to show that $|x_m-x_n|\leq\frac{1}{\sqrt{\min\{m,n\}}}$. So for any $\epsilon>0$, we can choose $N$ such that $N>\frac{1}{\epsilon^2}$, and then $|x_n-x_m|<\epsilon$ for any $n,m>N$.