Prove series $$\sum\limits_{n=2}^{\infty}\frac{1}{n\ln n}$$ doesn't converge without integral test. I really don't know any hint for $\ln n$. So I understand I have to show $$\frac{1}{n\ln n} > \frac{1}{x^b}$$ where $b<1$. But I really don't know how
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$\begingroup$This is really a classic example for the usefulness of the Cauchy condensation test, the rather surprising but easy to prove fact that a positive series $$ \sum_{n\in \mathbb{N}}a_n $$ converges if and only if the series $$ \sum_{k\in \mathbb{N}}a_{2^k}2^k $$ does as well. Applying this we may examine $$ \frac{1}{\log 2}\sum_{n\geq 2}\frac{1}{n} $$ which we know diverges.
$\endgroup$ 1 $\begingroup$we can use Cauchy condensation test if $(u_{n})_{n \in\mathbb{N}}$ is decreaisng and $\geq$ 0 then $\sum$$u_{n}$ converge iff $\sum$$2^{p} u_{2^p}$ converge
$\endgroup$ 1 $\begingroup$Hint:) Use Cauchy condensation test.
$\Big\{\dfrac{1}{n\ln n}\Big\}$ is a positive and decreasing sequence, then $$\sum\limits_{n=2}^{\infty}\frac{1}{n\ln n} \color{blue}{~\text{converges}} \Leftrightarrow \sum\limits_{n=2}^{\infty}2^n\frac{1}{2^n\ln 2^n} \color{blue}{~\text{converges}}$$
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