Is this sequence Cauchy?


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

  • $\begingroup$ Suppose that converges as it goes to 0 and is therefore Cauchy $\endgroup$ – Daniela del Carmen May 8 '12 at 4:28
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    $\begingroup$ 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 $0$". $\endgroup$ – Arturo Magidin May 8 '12 at 4:30
  • $\begingroup$ 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. $\endgroup$ – Arturo Magidin May 8 '12 at 4:30
  • $\begingroup$ Every convergent sequence is Cauchy, see e.g. proofwiki. $\endgroup$ – Martin Sleziak May 8 '12 at 4:39

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 $0$)

  • $\begingroup$ (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 $\endgroup$ – Daniela del Carmen May 8 '12 at 4:32
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    $\begingroup$ @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...) $\endgroup$ – Arturo Magidin May 8 '12 at 4:35

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