Question

How do you use the direct comparison test to determine if `Sigma 1/(3n^2+2)` from `[1,oo)` is convergent or divergent?

Answer 1

See below.

Explanation:

`3n^2+2 > 3n^2` so

`1/3sum_(k=1)^oo 1/n^2 > sum_(k=1)^oo 1/(3n^2+2)`

We know that `sum_(k=1)^oo 1/n^2` is convergent so `sum_(k=1)^oo 1/(3n^2+2)` also is