How is this limit equal to zero: $\lim_{n\to\infty}\frac{(\log n)^2}{n^\frac{1}{4}}=0$? The whole question is to prove that $$y=\frac{(\log n)^2}{n^\frac{1}{4}} \text{ is a convergent series}.$$
In his answer, the author has said that $$\lim_{n\to\infty}\frac{(\log n)^2}{n^\frac{1}{4}}=0 \text { and } (\log n)^2<n^{\frac{1}{4}}. $$
Can someone please explain how exactly did the author conclude that limit point?
I understood the other one as $\lim_{n\to \infty}y$ is somewhere $<1$ by plotting a graph. 
 A: Note that $\frac{(\ln n)^2}{n^\frac{1}{4}}=\left(\frac{\ln n}{n^\frac{1}{8}}\right)^2$, so it suffices to consider, for $a>0$, the limit
$$\lim_{x\to+\infty}\frac{\ln (x)}{x^a}=\lim_{y\to+\infty}\frac{y}{e^{ay}}=\lim_{y\to+\infty}\frac{1}{ae^{ay}}=0$$
where we set $y=\ln(x)$ in the last step we used Hopital's rule.
A: Even $$\lim_{n\rightarrow+\infty}\frac{(\ln{n})^{10000000000000}}{n^{0.000000000001}}=0$$
A: It suffices to show the log of the expression tends to $-\infty$:
$$\log\Biggl(\frac{\log^2n}{n^{\tfrac14}}\Biggr)=2\log(\log n)-\frac14\log  n=-\frac14\log  n\biggl(1-8\frac{\log(\log n)}{\log n}\biggr).$$
Now, by substitution, $\dfrac{\log(\log n)}{\log n}\to 0$ as $n\to\infty$, so the second factor tends to $1$, and the log of the expression has the limit of the first factor.
Concerning the series, this is a divergent Bertrand's series. These are series of the form
$$\sum_{n}\frac1{n^r \log^sn},\qquad (\text{here}\quad r=1/4,\;s=-2)$$
and they are known to converge if and only if\begin{cases}r>1\quad\text{or}\\ r=1\enspace\text{and}\enspace s>1.\end{cases}
A: It is a known fact that $\log(n) < n^r$ for all $r>0$ and sufficiently large $n$. 
This yields that in fact $\lim_{n\to \infty} \frac{\log(n)}{n^t} = 0$ for all $t>0$ (apply the fact with any $r>t$). 
And hence $\lim_{n\to \infty} \frac{\log(n)}{n^{1/8}} = 0$, so $\lim_{n\to \infty} \frac{\log(n)^2}{n^{1/4}} = 0$.
However, this does not imply $y=\frac{(\log n)^2}{n^\frac{1}{4}} \text{ is convergent series}$. It is a necessary but not a sufficient condition (by the Divergence test) that $\lim_{n\to\infty}\frac{(\log n)^2}{n^{1/4}}=0$.
In fact, the series is divergent by comparison to $\frac1n$. 
