How do you use the direct Comparison test on the infinite series #sum_(n=1)^ooln(n)/n# ?

1 Answer
Wataru
Sep 24, 2014

By Comparison Test with the harmonic series,

#sum_{n=1}^infty{ln n}/n# diverges.

Let us look at some details.

For #n ge 3#,

#ln n ge ln 3 ge 1 Rightarrow {ln n}/n ge 1/n#

(Note: It is okay even if the first few terms do not satisfy the inequality since the sum of those terms is finite, which does not affect the convergence of the whole series.)

Since #sum_{n=1}^infty 1/n# is a divergent harmonic series, we conclude that

#sum_{n=1}^infty{ln n}/n# also diverges by Comparison Test.

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