Prove that if there exists a constant $a$ such that $0<a\le x_n$ then $$\lim\sup{1\over x_n}={1\over \lim\inf x_n}$$. $\{x_n\}$ is bounded.

It is important for me to know where exactly I am wrong. Here is my attempt:

$x_n$ is lower bounded and therefore must have a minimal partial limit, $m=\lim \inf x_n$ (I don't know how to explain why formally)

$x_n$ is upper bound then it must have a maximal partial limit (I don't know how to explain why formally), $s=\lim \sup x_n$. Then $\lim\sup {1\over x_n}=\max PL\{{1\over m},...,{1\over s}\}={1\over m}={1\over \min PL\{m,...,s\}}={1\over \lim\inf x_n}$.

Why do I actually need $a$? I would appreciate your reply.

  • 1
    $\begingroup$ Because $\frac{1}{0}$ is not well defined I presume (take $x_n = \frac{1}{n}$) $\endgroup$
    – Tryss
    Feb 25, 2015 at 0:04

1 Answer 1


Since $x_n \ge a > 0$ for all $n$, $\liminf x_n \ge a > 0$, hence $1/\liminf x_n$ is a finite number. Now since $\frac{1}{x_n} \le \frac{1}{a}$, $\limsup \frac{1}{x_n} \le \frac{1}{a} < \infty$ So the equation above at least makes sense. Without having that $a$ present in the hypothesis, you can have a $1/0$ situtation (e.g., if $x_n = 1/2^n$), which is problematic.

To show that the equation holds, let $L = \limsup 1/x_n$ and $M = \liminf x_n$. Given $\epsilon > 0$, $1/x_n < L + \epsilon$ for all but finitely many $n$. Thus

$$x_n > \frac{1}{L + \epsilon}$$

for all but finitely many $n$. Hence

$$\liminf x_n \ge \frac{1}{L + \epsilon},$$

that is, $M \ge 1/(L + \epsilon)$. On the other hand, by definition of $M$, $x_n > M - \epsilon$ for all but finitely many $n$. Thus

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

for all but finitely many $n$. Therefore

$$\limsup \frac{1}{x_n} \le \frac{1}{M - \epsilon},$$

or $L \le \frac{1}{M - \epsilon}$. Letting $\epsilon \to 0^+$ results in $M \ge 1/L$ and $L \le 1/M$. In other words, $L \ge 1/M$ and $L \le 1/M$. Therefore $L = 1/M$, as desired.

  • $\begingroup$ If $a_n<L+\epsilon$ (for example) for all but finitely many number of elements, doesn't it mean that $L$ is the only limit? $\endgroup$ Feb 25, 2015 at 6:28
  • $\begingroup$ @MeitarAbarbanel it implies that $\limsup a_n \le L + \epsilon$. $\endgroup$
    – kobe
    Feb 25, 2015 at 6:31

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