Question Does $\sum^\infty_{n=3}\frac{1}{n\ln n}$ converge or diverge?
Attempt. I went with the fact that $\ln n<\sqrt n$ for all $n\geq0$. And proceeded to attempt the direct comparison test: $$\sum^\infty_{n=3}\frac{1}{n\ln n}\leq \sum^\infty_{n=3}\frac{1}{n\sqrt n}=\sum^\infty_{n=3}\frac{1}{n^{\frac32}}$$ which by the p-series, $\frac32\geq 1$ hence the right part converges, so the $\sum^\infty_{n=3}\frac{1}{n\ln n}$ should as well, but I've seen this sum to be divergent by the integral method. Where is my mistake?