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Is this sequence Cauchy?

$\left\{\frac{1}{n^{2}}\right\}$

My attempt: Suppose that converges as it goes to $0$ and is therefore Cauchy, but I lack formality in my reply.

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    Every co$n$vergent sequence is Cauchy, see e.g. [proofwiki](http://www.proofwiki.org/wiki/Convergent_Sequence_is_Cauchy_Sequence).2012-05-08

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In order to be Cauchy, it must be the case that for all $\epsilon\gt 0$ there exists $N\gt 0$ such that, for all $n,m\geq N$, we have $\left|\frac{1}{n^2}-\frac{1}{m^2}\right|\lt \epsilon.$ Let us assume without loss of generality that $n\geq m$. Then $\left|\frac{1}{n^2}-\frac{1}{m^2}\right| = \frac{1}{n^2}-\frac{1}{m^2} \lt \frac{1}{n^2}.$ If we can ensure that $\frac{1}{n^2}$ is small enough, provided $n$ is large enough, then that would suffice. Can we?

Added. Well, if we want $\frac{1}{n^2}\lt \epsilon$, then, since both $\epsilon$ and $n$ are positive, we need $\frac{1}{\epsilon}\lt n^2,$ which means we need $\frac{1}{\sqrt{\epsilon}}\lt n.$ So... what is a good $N$ to pick so that, if $n\geq N$, then $\frac{1}{n^2}\lt\epsilon$?

(What is behind this particular estimate is that: (i) every convergent sequence is Cauchy; and (ii) this sequence converges to $0$)

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    @DanieladelCarmen: Which amounts to showing that if $\epsilon\gt0$, then there exists $N\gt 0$ such that for all $n\geq N$, $\frac{1}{n^2}\lt \epsilon$; that is, I've reduced the problem of showing that it is Cauchy to showing that it converges to $0$. At this point, can you not see how to ensure that $\frac{1}{n^2}\lt\epsilon$ by asking that $n$ be large enough? How large should $n$ be so that $\frac{1}{n^2}\lt\epsilon$ holds? (I did not know you had already thought of that, because *at the time I wrote the answer* you had not yet bothered to tell us what you had already thought of...)2012-05-08