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How to formally prove that if $\lim \limits_{n \to \infty}a_n=\infty$, then $\lim \limits_{n \to \infty}\frac{1}{a_n}=0$.

I got confused for how to mix between convergence to infinity and having finite limit. how do I write a proof to link between two things? ( by using $\epsilon $) and not just giving an example.

my proof:

let $M= \frac{1}{\epsilon}$. there exists such a $N\in\mathbb N$ s.t for every $n>N$, it is true that: $$a_n>\frac{1}{\epsilon}$$

we should prove that for every $\epsilon>0$ there exists such a $N\in\mathbb N$, s.t for every $n>N$ $$\left|\frac1{a_n}\right|<\epsilon$$

so we take the N that satisfies the first conclusion, and that will mean for every $n>N$$$\left|\frac1{a_n}\right|<\epsilon$$

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1 Answer 1

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Use the strict definition. That is, use the fact that:

For every $M\in \mathbb R$, there exists such a $N\in\mathbb N$ that for every $n>N$, it is true that $$a_n>M$$

And prove the fact that:

For every $\epsilon\in \mathbb R$, there exists such a $N\in\mathbb N$ that for every $n>N$, it is true that $$\left|\frac{1}{a_n}\right|<\epsilon$$

To do this:

  1. Take an arbitrary $\epsilon$.
  2. Try to see how large the value of $a_n$ must be in order for the value of $\left|\frac{1}{a_n}\right|$ to be below $\epsilon$. Call that value $C$.
  3. You can now find the value of $N$ for which $a_n>C$ for all $n>N$, meaning that $\left|\frac1{a_n}\right|<\epsilon$ for all $n>N$.
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  • $\begingroup$ can I take $M= \frac{1}{\epsilon}$ according to the first definition?. and because after certain $n$, $a_n >0$, and I can take it out of the absolute value and find the relation between $M$ and $\epsilon$. $\endgroup$
    – Xhero39
    Feb 16, 2015 at 10:22
  • $\begingroup$ @FirasAliAbdelGhani I think you have the correct idea, though the way you phrased it is a little awkward. If you write your proof into the question by editing it, I can tell you exactly if it is ok. $\endgroup$
    – 5xum
    Feb 16, 2015 at 10:27
  • $\begingroup$ I edited. what do you think ? $\endgroup$
    – Xhero39
    Feb 16, 2015 at 10:54
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    $\begingroup$ @FirasAliAbdelGhani Looks fine to me. Just add 2 things: First of all, before saying "let $M=1/\epsilon$", you need to select the value of $\epsilon$. It is better to start your proof with "let $\epsilon > 0$" and go from there. And second, while it is simple, you have not pointed out the link between the fact that $a_n>1\epsilon$ and the fact that $\left|\frac1{a_n}\right|<\epsilon$. I think you know these things, but without writing them down, your professor may not know that you do. $\endgroup$
    – 5xum
    Feb 16, 2015 at 11:17
  • $\begingroup$ I think that you're totally right about that. thanks for helping!! $\endgroup$
    – Xhero39
    Feb 16, 2015 at 11:40

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