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.

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 $\sum f(n) \leq \sum 2^n f(2^n) \leq 2 \sum f(n)$.
