How do you use the direct Comparison test on the infinite series #sum_(n=1)^ooarctan(n)/(n^1.2)# ?

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Answer 1 · 2014-10-15T17:50:47

Since

#arctann le pi/2#,

we have

#{arctann}/n^{1.2} le {pi/2}/n^{1.2}#.

Since
#sum_{n=1}^infty {pi/2}/n^{1.2}=pi/2 sum_{n=1}^infty 1/n^{1.2}#

is a convergent p-series with #p=1.2 > 1#,

by Comparison Test,

#sum_{n=1}^infty {arctann}/n^{1.2}# also converges.

I hope that this was helpful.