How to show if #sum_(n=2)^oo 1/(n^2+ln n)# series converges or diverges by using comparison test?

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

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Answer 1 · 2018-03-27T10:02:12

The series:

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

is convergent.

For #n >= 2# we have that:

#lnn > 0#

#n^2+lnn > n^2#

#1/(n^2+lnn) < 1/n^2#

and as the series:

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

is convergent based on the #p# series test, then also:

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

is convergent.