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We want to show that $\lim_{n\to\infty} a_n = \infty \iff \lim_{n\to\infty} \frac{1}{a_n} = 0$, where $a_n > 0$ is a sequence, $n \in \mathbb{N}$.

My attempt at a solution: negate definition of convergence and work from there?

Suppose that $\lim_{n\to\infty} a_n = \infty$. Then there exists $\epsilon > 0$ such that for all $n \in \mathbb{N}$ whenever $n \geq N$, $|a_n| \geq \epsilon$. Then consider $|\frac{1}{a_n}|$. Let $n \geq N$. Then:
$|\frac{1}{a_n}| \leq |\frac{1}{a_N}| \leq |\frac{1}{\epsilon}| < \epsilon$, so that $\frac{1}{a_n}$ has limit 0.

Does that work?

Now with the other direction...
Suppose that $\lim_{n\to\infty} \frac{1}{a_n} = 0$. Hence there exists $N \in \mathbb{N}$ such that whenever $n \geq N$, $|\frac{1}{a_n}| < \epsilon$.
Let $n \geq N$. Then $|a_n| > \epsilon$.

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  • $\begingroup$ the last line, I think $|1/a_n|<\epsilon$ implies $|a_n|>1/\epsilon$. $\endgroup$ – Pei-Lun Tseng Apr 5 '17 at 4:55
  • $\begingroup$ What if I choose $\epsilon = \frac{1}{\epsilon}$? (Since we are proving divergence, I can pick epsilon, right?) $\endgroup$ – snailshell Apr 5 '17 at 5:00
  • $\begingroup$ $a_n=-1/n\to 0$, but $1/a_n =-n\to -\infty \ne \infty$ $\endgroup$ – Mark Viola Apr 5 '17 at 5:02
  • $\begingroup$ Oops. I forgot to add one of the conditions is that $a_n$ is positive for all $n$. $\endgroup$ – snailshell Apr 5 '17 at 5:03
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I think you got the idea but here's a revised version of your first proof to make the logic clearer; use $M$ to denote "big" numbers and $\epsilon$ for smaller ones.

Let $\epsilon > 0$. Pick $M > 0$ such that $\frac{1}{M} < \epsilon$. Since $a_n \rightarrow \infty$, there exists an $N\in\mathbb{N}$ such that $$n \geq N \implies a_n > M$$ or equivalently, $$n \geq N \implies \epsilon > \frac{1}{M} > \frac{1}{a_n}. $$

Then for the other direction, you want to "fix" a number $M>0$, and show that you can pick an $N$ such that $a_n > M$ for all $n \geq N$.

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