Question

How do you use the integral test to determine if `1/4+2/7+3/12+...+n/(n^2+3)+...` is convergent or divergent?

Answer 1

The series `sum_(n=1)^(+oo) n/(n^2+3)` is divergent.

Explanation:

The integral tests states that a necessary and sufficient condition for:

`sum_(n=1)^(+oo) a_n`

to converge is that the integral:

`int_1^(+oo) f(x)dx`

is convergent, where `f(x)` is chosen such that `f(n) = a_n`

If we pose:

`f(x) = x/(x^2+3)`

we can calculate:

`int_1^(+oo) x/(x^2+3)dx= 1/2int_1^(+oo) (d(x^2+3))/(x^2+3)=[1/2ln(x^2+3)]_1^(+oo)`

This integral is not convergent, and therefore the series is divergent.