I wish to prove the statement.
If $\lim_{n\to\infty}{\frac{s_n}{n}} >0$, then $s_n\to\infty$.
I will begin by assuming that $\lim_{n\to\infty}{\frac{s_n}{n}}=L$. Then I know that given $\epsilon>0$, $\exists N\in \mathbb{N}$ such that $|\frac{s_n}{n}-L|<\epsilon$ for all $n\geq N$.
Im not sure where to go from here. Any hints?
Let $L:=\lim_{n\rightarrow\infty}\dfrac{s_{n}}{n}>0$, then for some $N$, we have $\left|\dfrac{s_{n}}{n}-L\right|<\dfrac{L}{2}$ for all $n\geq N$, so $\dfrac{s_{n}}{n}>L-\dfrac{L}{2}=\dfrac{L}{2}$ for all such $n$, then given any $M>0$, for all $n\geq\dfrac{2M}{L}+N$, then $s_{n}=\dfrac{s_{n}}{n}\cdot n>\dfrac{L}{2}\cdot n>\dfrac{L}{2}\cdot\dfrac{2M}{L}=M$, this shows that $\lim_{n\rightarrow\infty}s_{n}=\infty$.