Having trouble with a series convergence-divergence test

Does the below series converge or diverge? $$\sum_{n=3}^{\infty}\dfrac{1}{n\ln (n-1)} $$ I have tried the integral test but stuck to either calculate that integral(substitution, integration by parts) or find an upper/below series that converges/diverges. Sorry for my bad English and Thanks in advanced!

Solution 1:

The given series is clearly comparable with the sum $$\sum_n \frac{1}{n\log{n}},$$ and you can compare this last one with the harmonic series $\sum_n \frac{1}{n}$ by using the Cauchy condensation test, which basically says that a series $\sum_n a_n$ of non-increasing terms $a_n$ will converge if and only if $\sum_n 2^n a_{2^n}$ converges.

Applying this comparison principle to $a_n=\frac{1}{n\log{n}}$, you deduce that the series diverges, as the harmonic series does.