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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.

Martin Sleziak
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The Dead Legend
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    $y=\frac{\log^2 n}{n^{1/4}}$ is not a series. $$\sum_{n\geq 1}\frac{\log^2 n}{n^{1/4}}$$ is a series, but it is not convergent. – Jack D'Aurizio Aug 20 '17 at 15:12

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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.

Robert Z
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    This limit is really a good example for students why they should not use the calculator for guessing a limit :). – MrYouMath Aug 20 '17 at 09:27
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Even $$\lim_{n\rightarrow+\infty}\frac{(\ln{n})^{10000000000000}}{n^{0.000000000001}}=0$$

Michael Rozenberg
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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}

Bernard
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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 $\frac1n$.

wythagoras
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    The answer seems to be circular. As you are assuming something that you want to show. – MrYouMath Aug 20 '17 at 09:25
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    $n<n+1$ for all $n $ but $\lim_{n\to\infty}\frac{n}{n+1} =1$ – Vincenzo Oliva Aug 20 '17 at 09:34
  • @MrYouMath The first and second fact are well-known, but can be proved using the Taylor series for $e^x$. – wythagoras Aug 20 '17 at 16:16
  • @VincenzoOliva The first fact is applied with any $r>t$, so this would be analogous to $n < (n+1) n^{\varepsilon}$ with $\varepsilon < 0$ and sufficiently large $n$, which isn't true. You are misunderstanding the argument (but note that I edited a bit to make it clearer) – wythagoras Aug 20 '17 at 16:18
  • The line that begins "This yields" really doesn't follow from the line above it, even after your edit. The point of @VincenzoOliva 's counterexample is that the inequality doesn't imply the limit. It is unclear how universally quantifying a parameter in the inequality is supposed to help. At the very least, there is still a gap in the proof. – John Coleman Aug 21 '17 at 01:39
  • Also, I don't think that you have fully responded to the claim of @MrYouMath that your answer seems to be circular. Of course it is a well-known fact, but so is the limit that you are trying to show. The most natural proof of the inequality is that it follows from the limit in question. Perhaps you can show the inequality from the Taylor series of $e^x$, but it isn't an immediate consequence of that series. Your appeal to a Taylor series is a bit of a hand-waive. – John Coleman Aug 21 '17 at 11:48