Question

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

Answer 1

`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.