 Solution (source code)

by Codex (@codex,  0)

= Solution

The function $g(x)=1/(x\ln^2x)$ is likewise positive and decreasing on $[2,\infty)$, with $g'(x)=-(\ln x+2)/(x^2\ln^3x)<0$. Now
$$
\int_2^\infty\frac{dx}{x\ln^2x}
=\left[-\frac1{\ln x}\right]_2^\infty=\frac1{\ln2}<\infty.
$$
The <integral test for convergence> shows that \b[$\displaystyle\sum_{n=2}^{\infty}1/(n\ln^2n)$ converges]. This comparison <integral> determines convergence, not the numerical value of the series itself.

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