How to prove that $\left\{\frac{1}{n^{2}}\right\}$ is Cauchy sequence

How can I prove that $$\left\{\frac{1}{n^{2}}\right\}$$ is a Cauchy sequence?

A sequence of real numbers $$\left\{x_{n}\right\}$$ is said to be Cauchy, if for every $$\varepsilon>0$$, there exists a positive integer $$N(\varepsilon)$$ such that $$\mid x_{n+p}-x_{n}\mid <\varepsilon$$ for all $$n\geq N$$ and $$p= 1, 2, 3,...$$

So I approached like this...

$$\mid \frac{1}{(n+p)^{2}}-\frac{1}{n^{2}}\mid = \frac{p(2n+p)}{n^{2}(n+p)^{2}}<\frac{p(2n+p)}{n^{2}} <\varepsilon$$

From here, I have to show that $$n>$$ some expression involving $$\varepsilon$$, because that expression will be the value of $$N$$. But I am getting stuck here.

Please anyone help me solve it. Thanks in advance.

• If a sequence is convergent, then it is Cauchy. (The converse is not necessarily true in non-complete spaces) – Julian Mejia May 25 at 16:27
• One of the simplest estimates may be $$\left|\frac1{(n+p)^2}-\frac1{n^2}\right|<\frac1{n^2}.$$ – Jyrki Lahtonen May 25 at 16:30
• @JulianMejia +1, Nothing more to say than this. – Michael Hoppe May 25 at 16:56

Since you already know what the limit is, this is not hard.

Let $$\epsilon > 0$$ be given. Choose $$N$$ such that $$\frac{1}{N^2} < \frac{\epsilon}{2}$$. Now assume $$n \ge N$$ and $$p \ge 1$$. Then $$|\frac{1}{n^2} - \frac{1}{(n+p)^2}| \le \frac{1}{n^2} + \frac{1}{(n+p)^2} \le \frac{2}{n^2} \le \frac{2}{N^2} < \epsilon \, .$$

• Ohh Thanks, Sir. Triangle inequality makes it so easy! – user587389 May 25 at 16:38

$$\frac{p(2n+p)}{n^2(n+p)^2}=\frac{p(2n+p)}{n^2(n^2+(2n+p)p)}\leq\frac{p(2n+p)}{n^2p(2n+p)}=\frac{1}{n^2}.$$

The triangular inequality is not needed, it is only application of $$0\le a\le b\implies 0\le b-a\le b$$

Let's have $$m\ge n$$ then apply to $$a=\frac 1{m^2}$$ and $$b=\frac 1{n^2}$$.

You get $$0\le \frac 1{n^2}-\frac 1{m^2}\le \frac 1{n^2}$$