Using limits to determine if $a_n=\frac{(\ln n)^2}{\sqrt{n}}$ (without L'Hospital) is convergent $a_n=\frac{(\ln n)^2}{\sqrt{n}}$ $(n\ge1)$
Would I use the squeeze theorem or can I just take the limit of top and bottom?
using L'Hopital, I get  $a_n=\frac{(4\ln n)}{\sqrt{n}}$ $(n\ge1)$

$$\ln{n}=4\ln{n^{\frac{1}{4}}}$$ and $$\sqrt{n}=(n^{\frac{1}{4}})^2$$ and you know $\lim_{x \rightarrow \infty} \frac{\ln{x}}{x}=0$ thus $$\lim_{n \rightarrow \infty}a_{n}=\frac{16(\ln{n^{1/4})^2}}{(n^{1/4})^2}=0$$
