The integral test may be applied to a sum #sum_(n=n_0)^oof(n)# if #f# is continuous, non-negative, and monotone decreasing on #[n_0,oo)#.
If that is the case, then #sum_(n=n_0)^oof(n)# converges or diverges together with the improper integral #int_(n_0)^oof(x)dx#
(Note that while these are the conditions typically used in the proof of the test, the integral test may actually be used under the much weaker condition that #f# is monotone on #[N,oo)# for some #N>=n_0#.)
a) We will use the test with the stronger, more typical conditions. As #ln(x)# and #x^2# are both continuous and positive on #[2,oo)#, we know that #f(x) = ln(x)/x^2# is continuous and non-negative on #[2,oo)#.
As #f'(x) = (1-2ln(x))/x^3<0# on #[2,oo)#, we know that #f(x)# is monotone decreasing on that interval. We can confirm the inequality by observing that for #x in [2,oo)#:
#x >= 2 > sqrt(e)#
#=> ln(x) > lnsqrt(e) = 1/2#
#=>1-2ln(x) < 1-2(1/2) = 0#
#=> (1-2ln(x))/x^3 < 0#
Thus, #f(x) = ln(x)/x^2# is continuous, non-negative, and monotone decreasing on #[2,oo)#, meaning the integral test may be applied to #sum_(n=2)^oof(n)#.
b) To apply the integral test, we'll first find the indefinite integral #intf(x)dx = intln(x)/x^2dx#.
Using integration by parts, let #u = ln(x)# and #dv = 1/x^2dx#. Then #du = 1/xdx# and #v = -1/x#, and so, by the integration by parts formula #intudv = uv - intvdu#, we have
#intln(x)/x^2dx = -ln(x)/x+int1/x^2dx = -ln(x)/x-1/x+C#
With that, we can find whether the improper integral #int_2^oof(x)dx# converges or diverges:
#int_2^oof(x)dx = lim_(N->oo)int_2^Nln(x)/x^2dx#
#=lim_(N->oo)((-ln(N)-1)/N-(-ln(2)-1)/2)#
#=0+(ln(2)+1)/2#
(verify that #lim_(N->oo)(ln(N)+1)/N=0# using L'Hopital's rule )
#=(ln(2)+1)/2#
Thus, as #int_2^oof(x)dx# converges, so does #sum_(n=2)^oof(n)# by the integral test.
