Does $\sum \limits_{n=1}^{\infty}{\frac{(\ln n)^2}{n^2}}$ converge?

Question

I'm trying to find out whether \begin{equation} \sum \limits_{n=1}^{\infty}{\dfrac{(\ln n)^2}{n^2}} \end{equation}

is convergent or divergent, but I can only use Comparison test or Ratio test.

I gave it some thought, but logarithms confused me. Regards and thanks a lot.

See also Show convergence/divergence for $\sum_{n=1}^{\infty} \frac{{(\ln n)}^{2}}{{n}^{2}} $ and Is this series convergent? $\sum_{i=1}^{\infty} \frac{(\log n)^2}{n^2}$. — Martin Sleziak, Dec 10 '18 at 18:43

score 10 · Answer 1 · answered Jul 01 '13 at 00:54

10

Since for all $\,n\,$ big enough we have that $\,\log n<n^\epsilon\;$ , with arbitrary $\,\epsilon>0\;$ , we get

$$\frac{\left(\log n\right)^2}{n^2}<\frac{\left(n^{1/4}\right)^2}{n^2}=\frac1{n^{3/2}}\ldots$$

answered Jul 01 '13 at 00:54

DonAntonio

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Thanks! But I'd like to prove $\log n<n^\epsilon$.Do you know an easy proof or a way to do that? – Ab urbe condita Jul 01 '13 at 02:45
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@user84523, what is the limit of $\frac{\log n}{n^\epsilon}$ when $n\to\infty$? You can use L'Hôpital's rule... – Mariano Suárez-Álvarez Jul 01 '13 at 06:26
See my answer, where I show $\ln x < x^\epsilon/\epsilon$ for any $\epsilon > 0$. – marty cohen Jul 02 '13 at 00:42

score 6 · Answer 2 · answered Jul 01 '13 at 00:46

Compare with $$\sum_{n=1}^{\infty} \dfrac1{n^s}$$where $s \in (1,2)$ and make use of the fact that $\dfrac{\ln(n)}{n^s} \to 0$ as $n \to \infty$ for $s > 0$. For instance, compare with $$\sum_{n=1}^{\infty} \dfrac1{n^{1.5}}$$

score 4 · Answer 3 · answered Jul 01 '13 at 21:54

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There's one more way. Compare it to the integral: $$ I=\int_{1}^{\infty}\bigg(\frac{\log x}{x}\bigg)^2dx $$ Also note $\frac{\log x}{x^2}<\big(\frac{\log x}{x}\big)^2<\frac{\log x}{x}<\frac{\log^2 x}{x}$and the limit of all these sequences is $0$ as $x \to \infty$. Now integrate I by parts in the following way: $$ I=\int_{0}^{\infty}1 \cdot \bigg(\frac{\log x}{x}\bigg)^2dx=\frac{\log^2 x}{x}\bigg|^{\infty}_{1}-2 \int_{1}^{\infty}\frac{\log x}{x^2}dx+2 \int_{1}^{\infty}\bigg(\frac{\log x}{x}\bigg)^2dx \\ =-2\int_{1}^{\infty}\frac{\log x}{x^2}dx+2I $$ so you get $$ I=2\int_{1}^{\infty}\frac{\log x}{x^2}dx $$ Again integrate the expression on RHS by parts to get $$ I=2(0+\int_{1}^{\infty}\frac{dx}{x^2})=2 $$ Hence the original sum converges too.

answered Jul 01 '13 at 21:54

Alex

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Thanks, I wish I could understand your solution (I'm gonna study integrals in two months). Anyway, I think could solve it using Cauchy's Condensation test: $\sum \limits_{n=1}^{\infty}{\dfrac{(\ln n)^2}{n^2}}$ converges iff $\sum \limits_{n=1}^{\infty}{\dfrac{2^n(\ln 2^n)^2}{(2^n)^2}}$, but $\dfrac{2^n(\ln 2^n)^2}{(2^n)^2}=\dfrac{(n\ln 2)^2}{(2^n)}<\dfrac{n^2}{2^n}$, and then use Ratio Test. Is it well done? – Ab urbe condita Jul 01 '13 at 22:26
That's pretty unusual to study sequence convergence, which are usually 2 year calculus, before integrals, which are high school stuff, but everyone is different. – Alex Jul 02 '13 at 04:35

score 2 · Accepted Answer · answered Jul 02 '13 at 00:12

To prove $\ln n < C(\epsilon) n^{\epsilon}$ where $ C(\epsilon)$ depends only on $\epsilon$:

Since $e^x \ge 1+x > x$, $x > \ln x$.

Now watch: The fingers never leave the hands!

Substitute $x^{\epsilon}$ for $x$. Then $x^{\epsilon} > \ln (x^{\epsilon}) = {\epsilon} \ln x $ so $\ln x < \dfrac{x^{\epsilon}}{{\epsilon}}$.

Setting ${\epsilon}=\frac{1}{m}$, $\ln x < m x^{1/m}$. In particular, for $m=2$, $\ln x < 2 x^{1/2}$.

Does $\sum \limits_{n=1}^{\infty}{\frac{(\ln n)^2}{n^2}}$ converge?

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