So the integral test states that, given a sum from some lower terminal to infinity, the integral with those terminals of what is being summed is defined, then the series if convergent.
To start, the function #f(n) = n^5e^(-n^2)# needs to be antidifferentiated, by repeated applications of the integration by parts formula:
#int n^5e^(-n^2) dn = -1/2 int n^4*-2n*e^(-n^2) dn#
(I write it like that for convenience in the next step).
Let #u_1 = n^4 rArr (du_1)/(dn) = 4n^3#
And #(dv)/(dn) = -2n*e^(-n^2) rArr v = e^(-n^2)# (this uses the 'inverse' version of the chain rule).
So, #-1/2 int u_1*(dv)/(dn) dn = -1/2 (u_1*v_1 - int v*(du_1)/(dn) dn)#
#=-1/2n^4e^(-n^2) + 1/2int 4n^3e^(-n^2) dn#
For now, let's set aside the #-1/2n^4e^(-n^2)#
Repeating the same process as before:
#1/2int 4n^3e^(-n^2) dn#
#= 2int n^3e^(-n^2) dn#
#=-int n^2*-2n*e^(-n^2) dn#
Let #u_2 = n^2 rArr (du_2)/(dn) = 2n#
And #(dv)/(dn) = -2n*e^(-n^2) rArr v = e^(-n^2)#
Again using the integration by parts formula:
#-int n^2*-2n*e^(-n^2) dn#
# = n^2e^(-n^2) - int 2n*e^(-n^2) dn #
Set aside the: #n^2e^(-n^2)#
Now all that's left is #- int 2n*e^(-n^2) dn#, which we know is just #e^(-n^2)# using the inverse chain rule as before.
So adding up all the little component parts:
#F(n) =int n^5e^(-n^2) dn#
#= -1/2n^4e^(-n^2) - n^2e^(-n^2) - e^(-n^2) + c#
#=-1/2e^(-n^2)(n^4 + 2n^2 + 2) + c#
Now on to the actual question! - evaluating the integral from one to infinity of this function, #f(n)#.
So the integral is...
#-1/2[e^(-n^2)(n^4 + 2n^2 + 2)]_1^oo#
Now substituting #n=1# into the antiderivative function, #F#, you get #F(1) = 5e^-1#
The improper part of the integral can be evaluated as the limit as #x# approaches infinity of #F(x)#, and this can be done as two limits because the limit of the product #e^(-n^2)*(n^4 + 2n^2 + 2)# is the same as the product of the limits.
So we want the #lim_(x rarr oo) F(x) - F(1)#
#=-1/2 (lim_(x rarr oo) e^(-x^2)*lim_(x rarr oo)(x^4 + 2x^2 + 2) - 5e^-1)#
#lim_(x rarr oo) e^(-x^2) = 0#
#lim_(x rarr oo)(x^4 + 2x^2 + 2) = oo#
So the overall integral is just... #5/2e^-1# - convergent!
Hope this helps!
