Is this series convergent? $\sum_{i=1}^{\infty} \frac{(\log n)^2}{n^2}$

Question

$\sum_{i=1}^{\infty} \frac{(\log n)^2}{n^2}$

I guess it is convergent, so I apply comparsion test for this.

$\frac{log^2}{n^2} < \frac{n^2}{n^2} = 1$

So it is bounded by 1 and hence it is convergent.

Can I do in this way? It kind of make sense, but I have never deal with constant in this case

Usually we do bounding using another term depend on $n$?

See also: Does $\sum \limits_{n=1}^{\infty}{\frac{(\ln n)^2}{n^2}}$ converge? and Show convergence/divergence for $\sum_{n=1}^{\infty} \frac{{(\ln n)}^{2}}{{n}^{2}} $ — Martin Sleziak, Dec 10 '18 at 18:45
Does this answer your question? Convergence of $\sum_{n=1}^\infty \frac{\ln^k(n) }{n^a}$ — Anne Bauval, Dec 04 '22 at 18:02

score 9 · Accepted Answer · answered May 30 '13 at 13:35

9

You can't do that way as Gerry mentioned. However, you know that $\log^2n\leq\sqrt n$ for $n$ big enough and that's why for big $n$ you have $\frac{(\log n)^2}{n^2}\leq \frac{1}{n\sqrt{n}}$. For the latter series do converge.

answered May 30 '13 at 13:35

SBF

36,568

You could make a mention why the latter series converges. – Vishal Gupta May 30 '13 at 13:48
@Vishal: I assume that the OP knows about the convergence properties of $\sum n^p$, true. Nevertheless, for the latter case the integral test can be used. – SBF May 30 '13 at 13:49
1

Mine looks so elementary among yours, so I preferred to remove that. +1 :-) – Mikasa May 30 '13 at 14:04
@BabakS.: I see, thanks – SBF May 30 '13 at 14:05
@BabakS. I would personnally do it the way you deleted...You lost an upvote. – Julien May 30 '13 at 15:16
@julien: I saw that I had nothing more than others to say. Thanks :) – Mikasa May 30 '13 at 19:04

score 4 · Answer 2 · answered May 30 '13 at 13:35

4

No, you can't do it that way. You have shown that the terms are bounded by $1$, but that's also true of the series $1+1+1+\dots$; does that series converge?

answered May 30 '13 at 13:35

Gerry Myerson

185,413

score 2 · Answer 3 · answered May 30 '13 at 13:38

What you did is wrong since you have to compare with the general term of a converging series.

You can try Cauchy's Condensation Test

$$\frac{2^nn^2\log^22}{2^{2n}}=\log^2\frac{n^2}{2^n}$$

and now it's easy to check that $\,\displaystyle{\sum_{n=1}^\infty\frac{n^2}{2^n}}\;$ converges (for example, the n-th root test)

score 2 · Answer 4 · edited Apr 13 '17 at 12:21

Note that, by the integral test

$$ \int_{1}^{\infty} \frac{(\ln(x))^2}{x^2}dx = -{\frac { \left( \ln \left( x \right) \right) ^{2}}{x}}-2\,{\frac { \ln \left( x \right) }{x}}-\frac{2}{x} \Big|_{x=1}^{\infty} =2 . $$

the series converges.

Main result: Suppose $f$ is continuous, positive, decreasing function on $[1,\infty)$ and let $a_n=f(n)$. Then

$(a)$ if $\int_{1}^{\infty}f(x) dx$ is convergent, then $\sum_{n=1}^{\infty} a_n $ is convergent.

$(b)$ if $\int_{1}^{\infty}f(x) dx$ is divergent, then $\sum_{n=1}^{\infty} a_n $ is divergent.

score 2 · Answer 5 · answered May 30 '13 at 15:10

Since $\log n\le n-1$ for all $n$, we get $\frac{1}{4}\log n \le \sqrt[4]{n}-1\le \sqrt[4]{n}$ for all $n$. So $$\frac{\log^2 n}{n^2}\le \frac{16\sqrt{n}}{n^2}=\frac{16}{n\sqrt{n}}$$ and we know that $\sum \frac{16}{n\sqrt{n}}$ converges. So by comparison test, $\sum \frac{\log^2 n}{n^2}$ converges.

score 0 · Answer 6 · answered May 31 '13 at 05:50

0

Let $\displaystyle a_n = \frac{\log^2 n}{n^2}$. Then $$\lim_{n\to\infty} n\left(\frac{a_{n+1}}{a_n}-1\right) = -2 < -1.$$ The series converges by Raabe's test.

answered May 31 '13 at 05:50

user26872

20,221

Is this series convergent? $\sum_{i=1}^{\infty} \frac{(\log n)^2}{n^2}$

6 Answers6

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