limsup and liminf are equal when...

Question

I have to prove the following:

If $\forall n\in \mathbb N \colon { \cfrac { 1 }{ 100 } <{ a }_{ n } }$ then $\limsup\limits_{ n\to\infty }\{ \frac { 1 }{ { a }_{ n } } \} =\frac { 1 }{ \liminf_{ n\to \infty }\{ { a }_{ n }\} } $.

Can anyone give me any pointers on this one?

Answer 1

Solved it (I think).

I showed that if a sequence $a_n$ has a positive lower bound, then ${ \sup\{ \frac { 1 }{ { a }_{ n } } \} }=\frac { 1 }{\inf{\{a_n}\}}$, and it follows that $\limsup=\liminf$.

Proof:

1. $\forall n\quad { a }_{ n }>\frac { 1 }{ 100 } \Rightarrow \exists i\in \Re \quad s.t.\quad i=inf\{ { a }_{ n }\} \quad \quad$

2. ${ a }_{ n }\ge i\ge \frac { 1 }{ 100 } \Rightarrow 100\ge \frac { 1 }{ i } \ge \frac { 1 }{ { a }_{ n } } $ so the sequence $\{ \frac { 1 }{ { a }_{ n } } \} $ is bounded above by $\frac { 1 }{ inf\{ { a }_{ n }\} } $.

3. Let's say there exists a smaller upper bound: $\exists j\in \Re \quad s.t.\quad \frac { 1 }{ i } >\frac { 1 }{ j } \ge \frac { 1 }{ { a }_{ n } } $.

4. It's all positive so this means: $i<j\le { a }_{ n } \Rightarrow i\neq inf\{ { a }_{ n }\} \bot $

5. So $\frac { 1 }{ \inf\{ { a }_{ n }\} }= \sup{\{\frac{1}{{a }_{n}}}\}$, and it's easy to see from here that: $\lim _{ n\rightarrow \infty }{ sup\left\{ \frac { 1 }{ { a }_{ n } } \right\} =\frac { 1 }{ \lim _{ n\rightarrow \infty }{ inf\{ } { a }_{ n }\} } } $