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

1 Answer
Dec 12, 2016

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.