Apply Cauchy's Condensation Test to ∑n=2∞1nlnn\sum_{n=2}^{\infty} \frac{1}{n \ln n}∑n=2∞nlnn1. The condensed series is:
∑k=1∞1k\sum_{k=1}^{\infty} \frac{1}{k}∑k=1∞k1 (constant multiple of harmonic series)
∑k=1∞2k2kln(2k)\sum_{k=1}^{\infty} \frac{2^k}{2^k \ln(2^k)}∑k=1∞2kln(2k)2k
∑k=1∞2kkln2\sum_{k=1}^{\infty} \frac{2^k}{k \ln 2}∑k=1∞kln22k
∑k=1∞1k2ln2\sum_{k=1}^{\infty} \frac{1}{k^2 \ln 2}∑k=1∞k2ln21