How do you use the direct comparison test to determine if #Sigma lnn/(n+1)# from #[2,oo)# is convergent or divergent?

1 Answer
Mar 5, 2017

We should first show that #sum_(n=2)^oo1/(n+1)# is divergent through limit comparison.

Letting #a_n=1/n# and #b_n=1/(n+1)# we see that

#lim_(nrarroo)a_n/b_n=lim_(nrarroo)(1/n)/(1/(n+1))=lim_(nrarroo)(n+1)/n=1#

Since #suma_n# is known to be divergent, we see that #sumb_n=sum_(n=2)^oo1/(n+1)# is divergent as well.

For #n>=3#, we see that #lnn>1#. Thus we can say that

#1/(n+1)<=lnn/(n+1)#

Since #1/(n+1)# is already a divergent series, the direct comparison test tells us that #sum_(n=2)^oolnn/(n+1)# is divergent as well.