# Why does this method of convergence fail? $\sum^\infty_{n=3}\frac{1}{n\ln n}$

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?

• If $\ln n<\sqrt n$ then $$\frac{1}{{\ln n}} > \frac{1}{{\sqrt n }}$$ is the right implication. You may use the integral test or the Cauchy condensation test to prove the divergence of your original series instead.
– Gary
Jul 11, 2021 at 18:04
• ahhh right, taking the reciprocal on both sides flips the inequality. Jul 11, 2021 at 18:17

Your error lies in assuming that $$\ln n<\sqrt n\implies\frac1{n\ln n}<\frac1{n\sqrt n}$$. Actually, it's the other way around.
Note the$$\int_3^\infty\frac{\mathrm dx}{x\ln x}=\lim_{M\to\infty}\ln(\ln M))-\ln(\ln(3))=\infty.$$Therefore, your series diverges.