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

1 Answer
Wataru
Apr 9, 2017

#sum tan(1/n)# is divergent.

Explanation:

Let #a_n= tan(1/n)# and #b_n=1/n#.

Let us check the hypotheses of Limit Comparison Theorem.

  1. #a_n>0# and #b_n>0# for all natural number #n#.

  2. #lim_(n to infty)a_n/b_n=lim_(n to infty)tan(1/n)/(1/n)#

By l'Hospital's Rule,

#=lim_(n to infty)(sec^2(1/n)cdot cancel(-1/n^2)) /cancel(-1/n^2)=sec^2(0)=1#

Since #sum b_n# is a harmonic series (divergent), by Limit Comparison Test, we can conclude that #sum tan(1/n)# is also divergent.

I hope that this was clear.

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