Although as , the divergence of the given series can be shown by both the Limit-Comparison and Integral tests (Table 8.3.1).
Since , which is negative for , the sequence is eventually monotone decreasing to zero, so the Integral test applies, and gives
=
Since this integral diverges, so too does the given series.
Alternatively, since = , by part (3) of the Limit-Comparison test the given series diverges because the comparison series is the divergent harmonic series.