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Let ${a_n}$ be a non-convergent (or divergent) sequence such that $1\le a_n \le 2$.

Prove or disprove:

$$\limsup a_n\cdot \limsup \frac 1 {a_n} \gt 1$$

My try:

$a_n$ must have at least two subsequential limits within the interval [1,2]. If that's the case then $\frac 1 {a_n}$ should have at least two subsequential limits within the interval $[\frac{1}{2}, 1]$. Thus, it follows that the inequality is true.

Is my approach correct?

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Your approach is pretty good, but I would note that since the limit does not exist, $$ \limsup_{n\to\infty}a_n\gt\liminf_{n\to\infty}a_n $$ and that $$ \limsup_{n\to\infty}\frac1{a_n}=\frac1{\liminf\limits_{n\to\infty}a_n} $$ Therefore, $$ \limsup_{n\to\infty}a_n\cdot\limsup_{n\to\infty}\frac1{a_n} =\frac{\limsup\limits_{n\to\infty}a_n}{\liminf\limits_{n\to\infty}a_n}\gt1 $$

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    $\begingroup$ Very elegant. +1 $\endgroup$
    – Fimpellizzeri
    Jan 29, 2018 at 20:16
  • $\begingroup$ Where can I find a formal proof of this claim: $\limsup_{n\to\infty}\frac1{a_n}=\frac1{\liminf\limits_{n\to\infty}a_n}$? $\endgroup$
    – Mister Bister
    Jan 29, 2018 at 20:17
  • $\begingroup$ @MisterBister: Suppose $a_n\gt0$. Then $$ \begin{align} \limsup_{n\to\infty}\frac1{a_n} &=\lim_{n\to\infty}\sup_{k\ge n}\frac1{a_k}\\ &=\lim_{n\to\infty}\lim_{m\to\infty}\max_{n\le k\le m}\frac1{a_k}\\ &=\lim_{n\to\infty}\lim_{m\to\infty}\frac1{\min\limits_{n\le k\le m}a_k}\\ &=\lim_{n\to\infty}\frac1{\lim\limits_{m\to\infty}\min\limits_{n\le k\le m}a_k}\\ &=\lim_{n\to\infty}\frac1{\inf\limits_{k\ge n}a_k}\\ &=\frac1{\lim\limits_{n\to\infty}\inf\limits_{k\ge n}a_k}\\ &=\frac1{\liminf\limits_{n\to\infty}a_n} \end{align} $$ $\endgroup$
    – robjohn ♦
    Jan 29, 2018 at 22:12

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