Solution
= Solution
Because $3>e$, each of $x$, $\log x$ and $\log\log x$ is positive and increasing for $x\ge3$. Hence $h(x)=1/[x\log x(\log\log x)^2]$ is positive, continuous and decreasing. With $u=\log\log x$, $du=dx/(x\log x)$, so
$$
\int_3^\infty\frac{dx}{x\log x(\log\log x)^2}=\int_{\log\log3}^\infty u^{-2}\,du=\frac1{\log\log3}<\infty.
$$
The <integral test for convergence> implies that the <series> is \b[convergent].