Solution
= Solution
The function $g(x)=1/(x\ln x)$ is positive and decreasing for $x\geq2$, since $g'(x)=-(\ln x+1)/(x^2\ln^2x)<0$. The <integral test for convergence> therefore applies. Its comparison <improper integral> is
$$
\int_2^R\frac{dx}{x\ln x}=\ln(\ln R)-\ln(\ln2)\longrightarrow+\infty.
$$
Consequently \b[$\displaystyle\sum_{n=2}^{\infty}1/(n\ln n)$ diverges].