# ${a_n}$ non-convergent sequence such that $1\le a_n \le 2$. Prove or disprove: $\limsup a_n\cdot \limsup \frac 1 {a_n} \gt 1$

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?

## 1 Answer

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, $$\limsup_{n\to\infty}a_n\cdot\limsup_{n\to\infty}\frac1{a_n} =\frac{\limsup\limits_{n\to\infty}a_n}{\liminf\limits_{n\to\infty}a_n}\gt1$$

• Very elegant. +1 – Fimpellizieri Jan 29 '18 at 20:16
• Where can I find a formal proof of this claim: $\limsup_{n\to\infty}\frac1{a_n}=\frac1{\liminf\limits_{n\to\infty}a_n}$? – Mister Bister Jan 29 '18 at 20:17
• @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} – robjohn Jan 29 '18 at 22:12