Question

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

Answer 1

Diverges.

Explanation:

`1/sqrt(n^3+1) > 1/sqrt(2n^3) = 1/sqrt(2) 1/n^(3/2)`

then

`1/sqrt(2) sum 1/n^(3/2) < sum 1/(sqrt(n^3+1))` and

`sum 1/n^(3/2)` diverges because `3/2 < 2` then

`sum 1/(sqrt(n^3+1))` diverges