Proving $\limsup\frac 1 {a_n}=\frac 1 {\liminf a_n}$ and $\limsup a_n\cdot \limsup \frac 1 {a_n} \ge 1$ 
Let $a_n$ be a sequence such that $\forall n\in \mathbb n: 0<a\le a_n\le b <\infty.$
Prove: 
  
  
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*$\displaystyle\limsup_{n\to\infty}\frac 1 {a_n}=\frac 1 {\displaystyle\liminf_{n\to\infty}a_n}$
  
*$\displaystyle\limsup_{n\to\infty}a_n\cdot \limsup_{n\to\infty}\frac 1 {a_n} \ge 1$ and there's an equality iff $a_n$ is converging.



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*Suppose there are two subsequences: $a_{n_l}, \ a_{n_k}$ such that $\lim a_{n_k} = k, \ \lim a_{n_l}=l$ and suppose $l\le k$, so $\lim \frac 1 {a_{n_k}}=\frac 1 k , \ \lim \frac 1 {a_{n_l}}=\frac 1 l$ so clearly: $\frac 1 k\le \frac 1 l\le l\le k$ so it's easy to see once the largest limit (supermum) is 'inverted' it has to become the smallest limit (infimum). 
I realize this doesn't show equality, I don't know how to do the other way and I'm not even sure if what I did is good.

*If $a_n$ converges, suppose to $L$ as its limit then we have: $L\cdot \frac 1 L=1$.
If it does not converge then $a_n$ may tend to infinity or won't have a limit. From 1 we can change it to $\displaystyle\limsup_{n\to\infty}a_n\cdot \frac 1 {\displaystyle\liminf_{n\to\infty}a_n} \ge 1$ and from BW, every sequence has a converging subsequence, and since for converging subsequences: $\liminf a_n\le \limsup a_n$ we have $\frac {\limsup a_n} {\liminf a_n}=\limsup a_n\cdot \limsup\frac 1 {a_n} \ge 1$. 
This should probably be in absolute value since one of those subsequnce limits can be negative, but it isn't in absolute value in the question. 

To get the first one, you just need to notice that
$$\sup\{1/{a_k}; k\ge n\} = \frac1{\inf\{a_k; k\ge n\}}$$
and take limit for $n\to\infty$ to get
$$\lim_{n\to\infty}\sup\{1/a_k; k\ge n\} = \lim_{n\to\infty}\frac1{\inf\{a_k; k\ge n\}} = \frac1{\lim_{n\to\infty}\inf\{a_k; k\ge n\}}.$$
(Although you should also check the cases when the LHS is $+\infty$ and when you have zero in the denominator.)
Once you have shown the first part, you get
$$\limsup_{n\to\infty} \frac1{a_n} = \frac1{\liminf_{n\to\infty} a_n} \ge \frac1{\limsup_{n\to\infty} a_n}.$$
Then you can simply multiply this by $\limsup\limits_{n\to\infty} a_n$.
Again, you should check separately the cases where some of the above values is $+\infty$ or when you divide by zero.
