How do you test the series #Sigma 1/(2+lnn)# from n is #[1,oo)# for convergence?

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Answer 1 · 2018-01-30T16:19:28

#sum_(n=1)^oo 1/(2+ln n) #

is divergent.

For #n >=2 # we have that #ln n < n#, so that:

#1/(2+ln n) > 1/(2+n)#

From the limit comparison test we can also see that as:

#lim_(n->oo) (1/(n+2)) / (1/n) = lim_(n->oo) n/(n+2) = 1#

and as:

#sum_(n=1)^oo 1/n #

is divergent, then also:

#sum_(n=1)^oo 1/(n+2) #

is divergent.

But then based on the direct comparison test also:

#sum_(n=1)^oo 1/(2+ln n) #

is divergent.