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

Question

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

Charles Arocha · Accepted Answer

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.

Isaiah Allen · Answer

undefined To use the Direct Comparison Test on the infinite series $ \sum_{n=1}^{\infty} \frac{\ln(n)}{n} $, we compare it with another series whose convergence properties we know. In this case, we can compare it with the harmonic series $ \sum_{n=1}^{\infty} \frac{1}{n} $. Since $ \ln(n) < n $ for all $ n > 1 $, we have:

$$ 0 < \frac{\ln(n)}{n} < \frac{1}{n} $$

Thus, by comparison, $ \sum_{n=1}^{\infty} \frac{\ln(n)}{n} $ is smaller than the harmonic series. Since the harmonic series diverges, by the Direct Comparison Test, $ \sum_{n=1}^{\infty} \frac{\ln(n)}{n} $ also diverges.