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 $|a_n| > \epsilon$.

  • $\begingroup$ the last line, I think $|1/a_n|<\epsilon$ implies $|a_n|>1/\epsilon$. $\endgroup$ 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

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 $n \geq N$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.