Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am attempting to prove if the below expression converges or diverges:

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

I decided to try using the limit convergence test. So then, I set $f(n) = \frac{1}{n (\ln n)^2}$ and $g(n) = \frac{1}{n}$ and did the following:

$$ \lim_{n \to \infty} \frac{f'(n)}{g'(n)} = \lim_{n \to \infty} \frac{-\frac{\ln(n) + 2}{n^2 (\ln n)^3}}{-\frac{1}{n^2}} = \lim_{n \to \infty} \frac{\ln(n) + 2}{n^2 (\ln n)^3} \cdot n^2 = \lim_{n \to \infty} \frac{1}{(\ln n)^2} + \frac{2}{(\ln n)^3} = 0 $$

I know that $\sum \frac{1}{n}$ diverges due to properties of harmonic series, and so concluded that my first expression $\frac{1}{n (\ln n)^2}$ also must diverge.

However, according to Wolfram Alpha, and as suggested by this math.SE question, the expression actually converges.

What did I do wrong?

share|improve this question
1  
You can use Cauchy's condensation test. –  Amihai Zivan Nov 18 '12 at 21:50
    
Your $f(x),g(x)$ do not depend on $x$, hence $f'(x)=g'(x)=0$. But you derive them $dn$, where $n$ is discrete. You should take much more care when doing things like that. –  Dennis Gulko Nov 18 '12 at 21:52
2  
The integral test works well too. –  mrf Nov 18 '12 at 21:55
    
is it $\log(x^2)$ or $(\log x)^2$? –  user31280 Nov 18 '12 at 22:04
    
@F'OlaYinka -- the latter –  Michael0x2a Nov 18 '12 at 22:10

4 Answers 4

up vote 7 down vote accepted

In the limit comparison test, $\sum_n f(n)$ and $\sum_n g(n)$ both will behave the same iff $$\lim_{n \to \infty} \dfrac{f(n)}{g(n)} = c \in (0,\infty)$$ In your case, the limit is $0$ and hence you cannot conclude anything.

share|improve this answer

You can use integral test for convergence. Since the integral $$\int_2^\infty\dfrac{1}{x\log^2 x}dx$$ converges, the series $\displaystyle{\sum_{n=2}^{\infty} \frac{1}{n \ln(n)^2}}$ converges.

share|improve this answer
1  
Thank you -- this is what I ended up using, but I decided to accept Marvis's answer because it helped show me why the limit test wasn't working here. –  Michael0x2a Nov 18 '12 at 22:12

Expanding on @Amihai Zivan's answer, you should use the Cauchy condensation test: For a positive non-increasing sequence $f(n)$, the series $\sum_{n=0}^\infty f(n)$ converges if and only if the series $\sum_{n=0}^\infty2^nf(2^n)$ does.
In your case, $f(n)=\frac{1}{n\ln^2n}$, so $$2^nf(2^n)=2^n\frac{1}{2^n\ln^2(2^n)}=\frac{1}{n^2\ln^22}=\frac{1}{\ln^22}\frac{1}{n^2}$$ Which clearly converges.

share|improve this answer

The numerical constant and further literature are in http://oeis.org/A115563 .

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.