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.
