Please help me to prove $\displaystyle\lim_{n\to\infty}a_n=0$ then $\displaystyle\lim_{n\to\infty}\frac{1}{a_n}=\infty$ Please give me a hint, not a full solution.

I know how to prove $a_n\to\infty \Rightarrow \frac{1}{a_n}\to0$, but not the other way around.

The original problem: Given $\forall a\in\left\{ a_n \right\}, a<0$ and $\displaystyle\lim_{n\to\infty}a_n=0$ prove: $\displaystyle\lim_{n\to\infty}\frac{1}{a_n}=-\infty$

  • 3
    $\begingroup$ Your statement is not true, consider $a_n=\frac{(-1)^n}{n}$. $\endgroup$ – iiivooo Mar 31 '16 at 18:38
  • $\begingroup$ If $\lim_{n\to\infty}a_n=0$, then $\lim_{n\to\infty}\frac{1}{a_n}$ exists if and only if $a_n$ has eventually constant sign. "Exists"here includes the possibility of the limit being infinite; presumably, no $a_n$ is zero. $\endgroup$ – MPW Mar 31 '16 at 18:49
  • $\begingroup$ What you wrote in "The original problem" must be part of the question from the very beginning: it makes all the difference ! $\endgroup$ – DonAntonio Mar 31 '16 at 20:04
  • $\begingroup$ @Joanpemo, agreed, but it is very important to me to not to get a full solution, but to get a general idea how to solve the problem. $\endgroup$ – Tegra Morgan Apr 1 '16 at 9:24

Since $\{a_n\}$ is negative and $a_n\to 0$, for each $M\in\mathbb{N}$ there exists $N$ such that $-\frac{1}{M}<a_n<0$ for all $n\geq N$.

Therefore $\frac{1}{a_n}<-M$ for all $n\geq N$, which implies that $\frac{1}{a_n}\to-\infty$.


Just which $\text{“}\infty\text{''}$ are we talking about? In some contexts it makes sense to distinguish between $+\infty$ and $-\infty$, and in some other contexts it makes sense to talk about just one $\text{“}\infty\text{''}$ that's at both ends of the real line and is approached by going in either the positive or the negative direction. For example, $\lim\limits_{x\to\pi/2} \tan x = \infty$, where this is the last-mentioned $\infty$. It is this last-mentioned $\infty$ that is approached by the reciprocal of something that approaches $0$.

If $a_n$ is close to $0$, then $-\varepsilon < a_n < \varepsilon$. That implies $a_n>1/\varepsilon$ or $a_n<-1/\varepsilon$.

To show $1/a_n$ approaches $\infty$, you need to show that no matter how big a number $N$ is, $1/a_n$ ultimately gets bigger than $N$ in absolute value, i.e. either bigger than $N$ or less than $-N$.

So let $\varepsilon = 1/N$.

However, if you want $1/a_n$ to approach $+\infty$ or $-\infty$, then you need $a_n$ to be always positive or always negative, respectively, with only finitely many exceptions.


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