Convergence or divergence of $\sum_{n=1}^{\infty} \frac{\ln(n)}{n^2}$

Question

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

The series is convergent or divergent? Would you like to test without the full ... I've thought of using the comparison test limit, but none worked, tried searching a number smaller or larger compared to use, but I got no ... Could anyone help me?

Please write correctly, because I am Brazilian and use translator ....

http://math.stackexchange.com/questions/425368/determine-whether-the-series-is-convergent-or-divergent — , Jun 20 '13 at 16:34

score 4 · Accepted Answer · answered Jun 20 '13 at 16:50

First proof" Cauchy's Condensation Test (why is it possible to use it?):

$$2^na_{2^n}=\frac{2^n\log(2^n)}{2^{2n}}=\frac{n}{2^n}\log 2$$

And now it's easy to check the rightmost term's series convergence, say by quotient rule:

$$\frac{n+1}{2^{n+1}}\frac{2^n}n=\frac12\frac{n+1}n\xrightarrow[n\to\infty]{}\frac12<1$$

Second proof: It's easy to check (for example, using l'Hospital with the corresponding function) that

$$\lim_{n\to\infty}\frac{\log n}{\sqrt n}=0\implies \log n\le\sqrt n\;\;\text{for almost every}\;\;n\implies$$

$$\frac{\log n}{n^2}\le\frac{\sqrt n}{n^2}=\frac1{n^{3/2}}$$

and the rightmost element's series converges ($\,p-$series with $\,p>1\,$) , so the comparison test gives us that our series also converges.

Also the integral test... – Adam Rubinson Oct 02 '22 at 12:33 — Adam Rubinson, Oct 02 '22 at 12:33

Git Gud · Answer 2 · 2013-06-20T16:42:47.720

3

Hint: Compare $x\to \sqrt x$ with $x\to \ln (x)$.

Further hint: The series $\sum \limits_{n=1}^{+\infty}\left(\frac{\sqrt n}{n^2}\right)$ converges.

edited Jun 20 '13 at 16:42

answered Jun 20 '13 at 16:33

Git Gud

31,356

1

@marcelo Feel free to ask for more hints in portuguese or whatever language you're more comfortable with. – Git Gud Jun 20 '13 at 16:35
Yes x is greater than sqrt (x), x is also larger than ln (x) So I can say that ? – benjamin_ee Jun 20 '13 at 16:40
@marcelolpjunior It is true that if $x>1$ the inequalities $x>\sqrt x$ and $x>\ln (x)$, but that doesn't help you. I'll add another hint. – Git Gud Jun 20 '13 at 16:41
It would not be possible to solve using comparison tests? With or without limit? – benjamin_ee Jun 20 '13 at 16:44
@marcelolpjunior That's exactly what I'm suggesting: compare $\sqrt n$ with $\ln (n)$. Do you get it? – Git Gud Jun 20 '13 at 16:46
Yes, just to confirm, we have that Σ n = 1, + ∞ (√ n / n ^ 2) is convergent and greater than ln (n) / n ^ 2, oh okay, just to show how even the series (√ n / n ^ 2) converge? Just to conluir – benjamin_ee Jun 20 '13 at 16:52
@marcelolpjunior To show the series in further hint converges, note it is a Dirichlet series, (check your notes). You should also make the observation that all the series involved are non-negative. Just to make sure you got it, how do you plan to prove that $\ln (n)\leq \sqrt n$? – Git Gud Jun 20 '13 at 16:55
Using the p series, it ... WOW, thanks a lot, it dawned on here lol ' – benjamin_ee Jun 20 '13 at 16:56
Great, thanks ... :) – benjamin_ee Jun 20 '13 at 16:57
@marcelolpjunior No problem. – Git Gud Jun 20 '13 at 16:59

Convergence or divergence of $\sum_{n=1}^{\infty} \frac{\ln(n)}{n^2}$

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