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.

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    $\begingroup$ Not at all. You are doing fine. $\endgroup$ – José Carlos Santos Nov 3 '18 at 16:40
  • $\begingroup$ Thank you professor @JoséCarlosSantos $\endgroup$ – Allorja Nov 3 '18 at 16:41
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    $\begingroup$ You've to replace $M$ by $M^2$, not $N$ by $N^2$. $\endgroup$ – Surajit Nov 3 '18 at 16:59
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    $\begingroup$ Tomath is right. Replacing $N$ with $N^2$ tells you nothing about how $\sqrt{s_n}$ compares with $M$. For instance, set $s_n=n\left(1+\frac{\sin n}{2}\right)$, $M=5$ and $N=5.96$. Even though $n>N\Rightarrow s_n>M$, we don't have $n>N^2\Rightarrow \sqrt{s_n}>M$. In fact, at $n=36>N^2$, we have $\sqrt{s_n}<M$. The right way to go about the proof is with $M\mapsto M^2$. $\endgroup$ – Jam Nov 3 '18 at 20:57
  • $\begingroup$ I believe you miss a crucial $\color{red}{+}$ sign, since we are allowed to say that $\lim_{n\to +\infty} n(-1)^n = \infty$ (meaning that $\lim_{n\to +\infty}\left|n(-1)^n\right|=+\infty$) but $\sqrt{s_n}$, in this case, is not always defined. $\endgroup$ – Jack D'Aurizio Nov 3 '18 at 21:32

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 $\lim_{n\to \infty}\sqrt s_n = \infty$


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.


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