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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.

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

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  • $\begingroup$ Ohh Thanks, Sir. Triangle inequality makes it so easy! $\endgroup$ – user587389 May 25 at 16:38
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$$\frac{p(2n+p)}{n^2(n+p)^2}=\frac{p(2n+p)}{n^2(n^2+(2n+p)p)}\leq\frac{p(2n+p)}{n^2p(2n+p)}=\frac{1}{n^2}.$$

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

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