多学科系统级建模仿真
 If it converges conditionally, determine if it also converge absolutely.
 Although an=lnn/n→0 as n→∞, the divergence of the given series can be shown by both the Limit-Comparison and Integral tests (Table 8.3.1).
 Since ddxlnxx=1−lnxx2, which is negative for x>e≐2.7, the sequence lnn/n is eventually monotone decreasing to zero, so the Integral test applies, and gives
 Since this integral diverges, so too does the given series.
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