Solution:
Assume an=nln(n)
limn→∞nlnn
=limn→∞11/n [Using L' Hopital rule]
=0
Next, Σan=∑n=1∞nln(n)
Consider ∫1∞andn=∫1∞nlnndn=[2(lnn)2]1∞=diverges
So, by integral series test, ∫1∞andn is divergent, then so is ∑n=1∞an
Thus, we have ∑n=1∞nlnn is divergent.
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