Question

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

Answer 1

See below.

Explanation:

We know that

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

Harmonic series. See https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

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.