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Let $c_n$ be a sequence that 'keeps away from zero', that is, $|c_n|≥r \,\forall n$ for some $r>0$ .

Let $a_n$ be a sequence that converges to 0, but $|a_n|>0$ for all $n$.

I have to prove that the sequence $\frac{c_n}{a_n}$ is divergent.

I tried assuming it was convergent and going through a proof for the sequence being cauchy, but I could not get a sensible contradiction (is that an oxymoron?).

Is it really as simple as saying $|\frac{c_n}{a_n}| ≥ |\frac{r}{a_n}|$ and then giving some simple argument, as to why this should be divergent?

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1 Answer 1

Yes. This is really this simple.

$$\left|\frac{c_n}{a_n}\right|\geq\left|\frac{r}{a_n}\right|=|r|\cdot\left|\frac1{a_n}\right|\to\infty$$

If you want a full $\varepsilon$-$\delta$ proof then one can extract it from the above argument:

Given $M>0$ we will find $n_0$ such that for all $n>n_0$ we have $M<\left|\frac{c_n}{a_n}\right|$. Because $a_n\to 0$ there exists some $n_0$ such that for $n>n_0$, $\left|a_n\right|<\frac rM$, therefore we have: $$\left|\frac{a_n}{c_n}\right|\leq\left|\frac{a_n}r\right|<\frac1M\implies \left|\frac{c_n}{a_n}\right|>M.$$

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@user49804: I didn't say that $\frac1{a_n}\to\infty$. I said that $\left|\frac1{a_n}\right|\to\infty$, because $|a_n|\to 0$. What is $t_n$ in your comment? I don't know. If it is supposed to be $c_n$, note that $c_n$ does not tend to $0$. –  Asaf Karagila Nov 17 '12 at 22:32
    
No, the comment locked up while I was editing it. In your proof as it stands now, you mean that $|\frac{1}{a_n}|\rightarrow \inf$ and not $|\frac{1}{a_n}|\rightarrow 0$, right? –  user49804 Nov 17 '12 at 22:39
    
@user49804: I see now what you were talking about. I corrected the argument. –  Asaf Karagila Nov 17 '12 at 22:39

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