The formal statement of the integral test states that if #fin[0,oo)rightarrowRR# a monotone decreasing function which is non-negative. Then the sum #sum_(n=0)^oof(n)# is convergent if and only if #"sup"_(N>0)int_0^Nf(x)dx# is finite. (Tau, Terence. Analysis I, second edition. Hindustan book agency. 2009).
This statement may seem a bit technical, but the idea is the following. Taking in this case the function #f(x)=xe^(-x)#, we note that for #x>1#, this function is decreasing. We can see this by taking the derivative. #f'(x)=e^(-x)-xe^(-x)=(1-x)e^(-x)<0#, since #x>1#, so #(1-x)<0# and #e^(-x)>0#.
Due to this, we note that for any #ninNN_(>=2)# and #x in[1,oo)# such that #x<=n# we have #f(x)>=f(n)#. Therefore #int_(n-1)^nf(x)dx>=int_(n-1)^nf(n)dx=f(n)#, so #sum_(n=1)^Nf(n)<=f(1)+sum_(n=2)^Nint_(n-1)^nf(x)dx=f(1)+int_1^Nf(x)dx#.
#int_1^oof(x)dx=int_1^ooxe^(-x)dx=-int_(x=1)^ooxde^(-x)=-xe^(-x)|_1^oo+int_1^ooe^(-x)dx##=-xe^(-x)-e^(-x)|^oo_1=2/e# using integration by parts and that #lim_(xrightarrowoo)e^-x=lim_(xrightarrowoo)xe^-x=0#.
Since #f(x)>=0#, we have #e/2=int_1^oof(x)dx>=int_1^Nf(x)dx#, so #sum_(n=1)^Nf(n)<=f(1)+2/e=3/e#. Since #f(n)>=0#, the series #sum_(n=1)^Nf(n)# increases as #N# increases. Since it is bounded by #3/e#, it must converge. Therefore #sum_(n=1)^oof(n)# converges.
