1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Because $\frac{1}{0}$ is not well defined I presume (take $x_n = \frac{1}{n}$) $\endgroup$ – Tryss Feb 25 '15 at 0:04
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Meitar Abarbanel Feb 25 '15 at 6:28
  • $\begingroup$ @MeitarAbarbanel it implies that $\limsup a_n \le L + \epsilon$. $\endgroup$ – kobe Feb 25 '15 at 6:31

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.