How do you use the limit comparison test to determine if $sum_(n=3)^(oo) 3/sqrt(n^2-4)$ is convergent or divergent?

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Answer 1 · 2017-12-17T18:10:29

See below.

We know that

$H = sum_(k=1)^oo 1/k = sum_(k=1)^oo a_k$ is divergent

Now considering

$S = sum_(k=3)^oo 3/sqrt(k^2-4) = 3 sum_(k=3)^oo b_k$

Here $a_k = 1/k < b_k = 1/sqrt(k^2-2)$ and $H$ diverges so $S$ diverges.

How do you use the limit comparison test to determine if sum_(n=3)^(oo) 3/sqrt(n^2-4)sum_(n=3)^(oo) 3/sqrt(n^2-4) is convergent or divergent?