I recommend theCondensation Test. If a_n is adecreasingsequence then Sum a_n converges or diverges with Sum 2^n a_{2^n}.

The proof is to bracket terms (a_1) + (a_2) + (a_3+a_4) + (a_5+...+a_8) + (a_9+...+a_16) + ... and compare.

In this case, condensation shows that Sum n^{-s} converges or diverges with Sum 2^n 2^{-ns} which converges if s>1 and diverges if s<=1. Applying it to Sum 1/(n log n) we consider Sum 2^n / (2^n n log 2) which we just saw diverges.

Another useful test in this case is theIntegral Test. Compare Sum f(n) with Int f(x) dx, assuming you can do the integral!

Finally, consider Sum k x^{k-1}. It looks like the derivative of something ...