Question

Find if series is convergent or divergent.

Series from n=2 to infinity
(4n+7)/(3n^3 -8n)

Answer 1

since 1/n^k converges for k>1, and

(4n+7)/(3n^3-8n) < (5n)/(3n^3) < 2/n^2

it converges.

Note that for sufficiently large n, 4n+7 < 5n.

And since 3n^3-8n < 3n^3, a smaller denominator makes for a larger quotient.