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!
