= Solution
The summand is positive, so it suffices to show that its <partial sums> are unbounded. On the block $2^k\leq n<2^{k+1}$, for $k\geq1$,
$$
\frac1{n\log n}\geq\frac1{2^{k+1}(k+1)\log2}.
$$
There are $2^k$ terms in that block, so its sum is at least $1/[2(k+1)\log2]$. Adding blocks gives a constant multiple of the divergent <harmonic series>. Hence
$$
\boxed{\sum_{n=2}^\infty\frac1{n\log n}=+\infty.}
$$
Equivalently the <integral test for convergence> gives $\int_2^R dx/(x\log x)=\log\log R-\log\log2\to\infty$. The block proof supplies the divergence without assuming the integral test.
Back to article page
