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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?

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    $\begingroup$ 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. $\endgroup$
    – Gary
    Jul 11, 2021 at 18:04
  • $\begingroup$ ahhh right, taking the reciprocal on both sides flips the inequality. $\endgroup$
    – Acyex
    Jul 11, 2021 at 18:17

1 Answer 1

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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.

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