# Prove $\lim\{s_n\}=+\infty\iff\lim\left(\frac{1}{\{s_n\}}\right)=0$

Prove the following: Given that $$\{s_n\}$$is a sequence of positive numbers. Then $$\lim s_n=+\infty\iff\lim\left(\frac{1}{s_n}\right)=0$$

My attempt at proving this:

For $$\lim s_n=+\infty\implies \lim\left(\dfrac{1}{s_n}\right)=0$$, suppose $$\lim s_n=+\infty.$$ Given any $$\epsilon>0$$, let $$K=\frac{1}{\epsilon}$$. Then there is a $$N\in\mathbb{N}\ni n \geq N\implies s_n>K=\frac{1}{\epsilon}$$. Since each of the $$s_n$$'s are positive, $$\mid \dfrac{1}{s_n}\mid<\epsilon$$, for $$n\geq N.$$ Therefore $$\lim(\dfrac{1}{s_n})=0.$$

Regarding the converse I'm having a bit more trouble; for $$\lim\left(\dfrac{1}{s_n}\right)=0\implies\lim s_n=+\infty$$.

Suppose that $$\lim \left(\dfrac{1}{s_n}\right)=0$$. Given any $$\epsilon>0$$, let $$M=\epsilon$$. Then there is $$N\in\mathbb{N}\ni n\geq N \implies |s_n-\infty|<\epsilon$$. Of course this makes no sense, as you can't subtract $$\infty$$, so I'm not sure how to proceed.

Any hints/tips would be appreciated.

What you have to show is the following: For every $K\in\mathbb{N}$ there is a $N\in\mathbb{N}$ such that $s_n\ge K$ for all $n\ge N$. This means that your sequence $\{s_n\}$ is converging to $+\infty$.
So let $K\in\mathbb{N}$ given. By assumption there is a $N\in\mathbb{N}$ such that for all $n\ge N$: $\frac{1}{s_n}\le\frac{1}{K}\iff K\le s_n$ for all $n\ge N$ and you are done.
Note, the last equivalence $\frac{1}{s_n}\le\frac{1}{K}\iff K\le s_n$ is the reason why your claim is true.
Hint: For the other direction start with: Let $N \in \mathbb{N}$ be arbitrary. Find some $\epsilon > 0$ such that $\epsilon < 1/N$. Now, since $1/s_n \to 0$ you can find an $M \in \mathbb{N}$ such that for every $n \geq M$, $1/s_n < \epsilon < 1/N$. Now, what can you conclude about $s_n$ for $n \geq M$?