$limsup{1over x_n}={1over liminf x_n}$?

Prove that if there exists a constant $a$ such that $0<ale x_n$ then $$limsup{1over x_n}={1over liminf x_n}$$. ${x_n}$ 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 $limsup {1over x_n}=max PL{{1over m},…,{1over s}}={1over m}={1over min PL{m,…,s}}={1over liminf x_n}$.

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

Answer

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.

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