If $\lim_{n\to \infty}s_n = \infty$ then also $\lim_{n\to \infty}\sqrt s_n = \infty$

Question

From the definition of diverging sequences we have that

$ \forall M>0 \ \exists N $ such that $n>N \implies s_n >M $

From the first limit, we know that such an N exists since we assumed that the limit diverges. Now if I take $N' = N^2$ for the limit I want to prove, will it suffice? If I am attacking this problem the wrong way, please suggest some other way.

Thank you.

Answer 1

We want to prove $\lim_{n\to \infty}\sqrt s_n = \infty$. Fix $M>0$ arbitrarily.
Then, as $\lim_{n\to \infty}s_n = \infty$, so there exists $n_0\in \mathbb{N}$ such that $s_n>M^2$ for all $n\geq n_0$, which gives $\sqrt s_n>M$ for all $n\geq n_0.$ Hence $\lim_{n\to \infty}\sqrt s_n = \infty$

Answer 2

If $\lim_{n\to \infty}\sqrt s_n = L<\infty$ then $\lim_{n\to \infty}(\sqrt s_n\cdot \sqrt s_n)$ exists and is equal to $(\lim_{n\to \infty}\sqrt s_n)\cdot (\lim_{n\to \infty}\sqrt s_n)=L^2$, which is a contradiction.