If $\lim_{n\rightarrow\infty} S_{n} = \infty$, prove that $\lim_{n\rightarrow\infty}\sqrt{S_{n}} = \infty$ I have no idea how to go about this and I have been stuck on it for 3 days.
The formula we have for infinite limits is:
For every $M>0$, there exists an $N \in \mathbb{N}$ such that for $n > N \implies S_{n} > M$.
The only thing that came to my mind is to replace $M$ with $K=M^{2}$, then take the square root of both sides and get $\sqrt{S_{n}}>M$. But I am not sure if its 'allowed'.

$\fbox{For every $M>0$, there exists an $N \in \mathbb{N}$ such that for $n > N \implies S_{n} > M$.}$
Let $M$ be arbitrary and choose $N \in \mathbb{N}$ such that for $n>N$, $S_n > M^2$. Then for $n>N$, $\sqrt{S_n} > M$

If you want to be really verbose, you could say: 
Let $M$ be arbitrary. By supposition, there exists an $N$ such that for $n>N$, $S_n > M^2$. Therefore, for $n>N$, $\sqrt{S_n} > M$, completing the proof.
