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

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

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2 Answers

up vote 4 down vote accepted

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.

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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.

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@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 ? –  marcelolpjunior Jun 20 '13 at 16:40
    
Thanks in advance :) –  marcelolpjunior Jun 20 '13 at 16:41
    
@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? –  marcelolpjunior Jun 20 '13 at 16:44
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