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How do I prove that $\displaystyle\sum_{n\geq 1}\frac {1}{\ln^2n}$ is a divergent series?

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What are the simplest cases of convergent and diveregence series and test that you know of? What have you tried? –  Arjang Aug 9 '11 at 12:48
    
I am not very good at math. I failed the exam. I tried to compare the series with others divergent series but I don't succeeded. Sorry for my english. –  NumLock Aug 9 '11 at 13:04
    
The first thing to ask is: does the bottom increase fast or slow? We know informally $\log n$ grows slowly. Its square is a lot faster, but still slow. For example, if we use logs to the base 2, $(\log_2(1024))^2=100$, well short of $1024$. So natural comparison is with $(1/n)$. –  André Nicolas Aug 9 '11 at 18:15

3 Answers 3

up vote 11 down vote accepted

Can you prove $\log n\lt\sqrt n$? Then $1/(\log n)^2\gt1/n$, and off you go.

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This is very easy to understand. Thank you too. –  NumLock Aug 9 '11 at 13:28

(Since I misinterpreted $\ln^2 n$ I rewrote the part specific to this question)

Suppose $\sum a_i$ and $\sum b_i$ are two infinite sums and $a_i\ge b_i\ge 0$ for all $i$.

We can you say if so that $\sum a_i\ge\sum b_i$, this is called a comparison test for convergence.

Suppose $\sum a_i$ is finite, then we have that $\sum b_i$ has to be finite as well. On the other hand, suppose $\sum b_i$ is infinite, in this case we have that $\sum a_i$ is infinite as well, simply by sandwiching: $$\lim_{n\to\infty}\sum_{i<n}b_i\le\lim_{n\to\infty}\sum_{i<n}a_i\le\infty$$

So in order to prove that $\sum\frac{1}{(\ln n)^2}$ is divergent, we need to find a sequence such that $a_n\ge(\ln n)^2$ and $\sum\frac{1}{a_n}$ diverges.

As Gerry suggests, $\ln n<\sqrt{n}$, which will lead to the required solution.

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But $\log\log n$ isn't $(\log n)^2$. –  Gerry Myerson Aug 9 '11 at 13:04
    
Thank you for your help –  NumLock Aug 9 '11 at 13:05
    
@Gerry: True dat, but $\ln^2 n$ can be read as $\ln\ln n$. –  Asaf Karagila Aug 9 '11 at 13:08
    
@Daniel: Please see my previous comment to Gerry, about the interpretation of $ln^2 n$. –  Asaf Karagila Aug 9 '11 at 13:09
    
Analytic number theorists occasionally need things like $\log\log\log\log n$, and to save space they may abbreviate it to $\log^4n$, so, yes, it can be read the way Asaf originally read it. But I think the default reading is the one NumLock actually meant. –  Gerry Myerson Aug 10 '11 at 0:42

Short answer: Since $n > \log n$ for all $n \in \mathbb R^+$, we know that $\frac{1}{n (\log n)} \leq \frac{1}{(\log n)^2}$ for all $n \in \mathbb R^+$.

We know that $\sum_{n=1}^\infty \frac{1}{n (\log n)}$ diverges, so by the comparison test, so does $\sum_{n=1}^\infty \frac{1}{(\log n)^2}$.

Longer answer: Here's some information that explains the motivation of comparing the series with $\sum_{n=1}^\infty \frac{1}{n (\log n)}$. It is a commonly known fact that all the following series diverge:

  • $\displaystyle \sum_{n=1}^\infty \frac{1}{n}$
  • $\displaystyle \sum_{n=1}^\infty \frac{1}{n (\log n)}$
  • $\displaystyle \sum_{n=1}^\infty \frac{1}{n (\log n) (\log \log n)}$
  • $\displaystyle \sum_{n=1}^\infty \frac{1}{n (\log n) (\log \log n)(\log \log \log n)}$
  • and so on

You can prove those by using the integral test (or the Cauchy condensation test, if you prefer). When you apply either of these tests, each series reduces to the series above it. We know the first one (which is the harmonic series) diverges, so by induction, all the other series diverge as well.

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I believe that $\frac{1}{\ln^2 n}=\frac{1}{\ln(\ln(n))}$. –  Asaf Karagila Aug 9 '11 at 12:54
    
BTW, still by the condensation test, the harmonic series reduces to the series $\sum 1$. –  Did Aug 9 '11 at 12:56
    
@Asaf: Hmm, that's a good point. If the original problem was with $\frac{1}{\log log n}$, I guess my solution was overkill. –  Alan C Aug 9 '11 at 13:01
    
@Didier: Oh yeah, thanks for pointing that out! –  Alan C Aug 9 '11 at 13:03
    
Thank you for your help Alan C . @ Asaf: In my country $ln^2n$ is $(log n)^2 $ , $log n $ natural logarithm. –  NumLock Aug 9 '11 at 13:05

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