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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

Thanks for your help

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What have you tried? –  Alex Becker May 8 '12 at 4:26
    
Suppose that converges as it goes to 0 and is therefore Cauchy –  Daniela del Carmen May 8 '12 at 4:28
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You should not suppose it converges, you should prove it converges. And it's not "as it goes to $0$"; it converges *as $n$ goes to $\infty$*, or else "it converges **to** $0$". –  Arturo Magidin May 8 '12 at 4:30
    
All-caps is the internet equivalent of shouting. You can use italics (enclosing text in * or in _) or bold (enclosing in **) if you want to convey emphasis. –  Arturo Magidin May 8 '12 at 4:30
    
Every convergent sequence is Cauchy, see e.g. proofwiki. –  Martin Sleziak May 8 '12 at 4:39
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1 Answer

up vote 2 down vote accepted

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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(i) every convergent sequence is Cauchy; and (ii) this sequence converges to 0),I had already thought of this but my problem was how to give a formal demonstration of this –  Daniela del Carmen May 8 '12 at 4:32
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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...) –  Arturo Magidin May 8 '12 at 4:35
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