Problem: Let $x_n \neq 0$ for all $n$ in the natural numbers. Prove that $|x_n| \rightarrow \infty$ if and only if $\frac{1}{|x_n|} \rightarrow 0$.

I've been noodling on this one for a long time, but can't seem to figure out how to dig my teeth in. Working on the direction of $\frac{1}{|x_n|} \rightarrow 0 \Rightarrow |x_n| \rightarrow \infty$, the best I've got is some algebra, trying to get a handle on $|x_n|$ in terms of $\frac{1}{|x_n|}$:

We know that, for all $\epsilon > 0$, we can find an $N$ in the natural numbers such that, for all $n > N$, we have $\Big|\frac{1}{|x_n|} - 0\Big| < \epsilon$.

So, for $n > N$, we have,

$-\epsilon < \frac{1}{|x_n|} - 0 < \epsilon$

$-\epsilon < \frac{1}{|x_n|} < \epsilon$

$-\frac{1}{\epsilon} > |x_n| > \frac{1}{\epsilon}$

I'm not sure, however, how this might help me show that $|x_n|$ is divergent. Am I barking up the wrong tree entirely? I'm hoping that, once I have a better idea how to prove the implication this way, I'll have some clearer thoughts about how to go the other way. Any help would be greatly appreciated.

Edit: Someone pointed out that I made a typo in the last inequality line.

  • $\begingroup$ Your last line is incorrect : when you take the reciprocal, you must switch the inequality, right? Also, note that $|x_n|$ is a positive quantity, so you need not focus on $|x_n| > \frac 1 \epsilon$. After switching the inequality, recall the definition of $x_n \to \infty$ to conclude one direction. $\endgroup$ – астон вілла олоф мэллбэрг Sep 9 '19 at 2:50
  • $\begingroup$ Since you are wrapping $x_n$ in absolute values, you can WLOG just prove the statement about nonnegative sequences and never have to deal with any absolute value signs at all. In other words, you only have to prove $x_n \to \infty$ iff $1/x_n \to 0$ for non negative sequences. $\endgroup$ – DanielV Sep 9 '19 at 3:18

Suppose $\frac{1}{|x_n|} \to 0$

Then, if $M \in \mathbb{R}_{>0}$, there is an positive number $N$ such that if $n > N$, then

$$\frac{1}{|x_n|} < \frac{1}{M}$$

But this implies that for such $n$, $M < |x_n|$. Since $M$ was an arbitrarily chosen positive real number, it follows that $|x_n| \to \infty$

For the converse, suppose $|x_n| \to \infty$. Then, fix $\epsilon > 0$. Then, there is a number $N>0$ such that if $n > N$, then $\frac{1}{\epsilon} < |x_n|$. But this implies $\frac{1}{|x_n|} < \epsilon$. The result follows.


What does it mean to say $|x_n|\rightarrow \infty$?

It means, given a real number $R$ of your choice, you can find $N\in \mathbb{N}$ (depending on $R$) such that $|x_n|>R$ for all $n\geq N$.

You have correctly realised that, given $\epsilon>0$, there exists $N\in \mathbb{N}$ such that $\frac{1}{|x_n|}<\epsilon$ (the next line you wrote is wrong). If you can correct it, you see that $|x_n|>\frac{1}{\epsilon}$. Do you see any relation with this observation and what I have said in above paragraph?


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