To make use of the integral test, we need to make sure that our function is decreasing, which we can check my taking a look at its derivative. we look at An = 1/(n^2-1) and then look at.
f(x) = 1/(x^2 -1) as we need to look at the function containing all real numbers, so that we can differentiate and integrate.
by using the Quotient rule we get:
f'(x) = -(2x)/(x^2-1)^2 which is always negative, meaning that our function is always decreasing.
we can also observe that the function is positive for all x >= 2, which is needed to use the integral test
now we know we can use the integral test for convergence.
so we evaluate the integral from 4 to oo as the question stated.
int_4^oo f(x) dx = int_4^oo 1/(x^2 -1) dx
=int_4^oo 1/((x+1)(x-1)) dx
where you can use Partial Fractions to get:
=int_4^oo (1/(2(x-1)) - 1/(2(x+1))) dx
= [1/2ln((x-1)/(x+1))] from x = 4 to oo
= lim x to oo [1/2ln((x-1)/(x+1))] - [1/2ln((4-1)/(4+1))]
= 1/2[ln(1) - (ln(3)-ln(5))]
= 1/2[ln(5) - ln(3)]
=1/2[ln(5/3)] ~~0.2554
Thus our integral converges from x = 4 to oo
and so we can conclude that our series sum 1/(n^2 -1) also converges due to the integral test for convergence.
