# Divergence of $\sum_{n = 2}^{\infty} \frac{1}{n \ln n}$ through the comparison test?

I have shown that it diverges through the integral test, but I am curious about how this would be shown using the comparison test. I can't use harmonic series because this is lesser than it. I had one idea: harmonic series can be compared to $1 + (\frac{1}{2}) + (\frac{1}{4} + \frac{1}{4}) + (\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8})$ to show that it diverges, maybe something similar can be done in this case?

Edit: using Cauchy condesnation:

$\sum_{n = 2}^{\infty} \frac{2^n}{2^n \log 2^n} \rightarrow \frac{1}{\log 2} \sum_{n = 2}^{\infty} \frac{1}{n}$, which is the harmonic series excluding $n = 1$, so the series diverges.

• See en.wikipedia.org/wiki/Cauchy_condensation_test Cauchy condensation test is basically comparison test – r9m Mar 4 '14 at 4:46
• Thank you. I have updated the post using this method. Please make sure that I am doing this correctly. – MCT Mar 4 '14 at 4:50
• Your first sentence says "that it converges" which is wrong. – Will Jagy Mar 4 '14 at 4:50
• @WillJagy Thanks, fixed (I know it diverges, just mis-typed.) – MCT Mar 4 '14 at 4:52
• Note that Cauchy Condensation is proved using a Comparison, so it exactly fulfills your request. – André Nicolas Mar 4 '14 at 4:55

Using Cauchy condensation, if $\displaystyle \sum_{n = 2}^{\infty} \frac{2^n}{2^n \log 2^n}$ converges or diverges, then the same must be true of my desired series.
This series is equal to $\frac{1}{\log 2} \displaystyle \sum_{n = 2}^{\infty} \frac{1}{n}$, the harmonic series, thus it diverges, and so does my desired series.
The comparison aspect of this series is inherent in the proof of Cauchy condensation. In particular, Cauchy condensation relies on the fact that $\sum f(n) \leq \sum 2^n f(2^n) \leq 2 \sum f(n)$.