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  $a_n=\frac{(4\ln n)}{\sqrt{n}}$ $(n\ge1)$
 A: Write the expression as:
$$a_n = \left(\frac{4\,\log n^{\frac{1}{4}}}{n^{\frac{1}{4}}}\right)^2$$
and you should make headway, using the well known fact that $log\,u/u\to0$ as $u\to\infty$.

You're simply using the definition of a continuous function in terms of limits for the function $f(x)=x^2$. To prove $\mathscr{L}=\lim\limits_{n\to\infty} \frac{\log n}{n}=0$ we make a substitution $n=e^u$ and then $\mathscr{L} = \lim\limits_{u\to\infty}\frac{u}{e^u}$. Since $e^u > 1 + u + \frac{u^2}{2!}$, this last one should be no problem.
A: $$\ln{n}=4\ln{n^{\frac{1}{4}}}$$ and $$\sqrt{n}=(n^{\frac{1}{4}})^2$$ and you know $\lim_{x \rightarrow \infty} \frac{\ln{x}}{x}=0$ thus $$\lim_{n \rightarrow \infty}a_{n}=\frac{16(\ln{n^{1/4})^2}}{(n^{1/4})^2}=0$$
A: Squeeze theorem is sufficient (and nice) here. More generally we prove that if $a, b$ are positive real numbers then $$\lim_{n \to \infty}\frac{(\log n)^{a}}{n^{b}} = 0\tag{1}$$ Let $f(n) = (\log n)^{a}/n^{b}$ and $c = b/a$ and $g(n) = (\log n)/n^{c}$ so that $f(n) = \{g(n)\}^{a}$ and if we show that $g(n) \to 0$ as $n \to \infty$ then $f(n) \to 0$ as $n \to \infty$.
We have $c > 0$ and hence it is possible to choose $d$ with $0 < d < c$. Let $n > 1$ so that $n^{d} > 1$ and then we know that $$0 < \log n^{d} < n^{d} - 1 < n^{d}$$ or $$0 < d \log n < n^{d}$$ or $$0 < \log n < \frac{n^{d}}{d}$$ and therefore $$0 < \frac{\log n}{n^{c}} < \frac{1}{dn^{d - c}}$$ Noting that $d - c > 0$ and applying Squeeze theorem when $n \to \infty$ we get $$\lim_{n \to \infty}\frac{\log n}{n^{c}} = 0\tag{2}$$ We have thus established that $g(n) \to 0$ as $n \to \infty$.
