Calculus of a Single Variable, Chapter 9, 9.3, Section 9.3, Problem 6

Integral test is applicable if f is positive and decreasing function on interval [k,oo) where a_n = f(x).
If int_k^oo f(x) dx is convergent then the series sum_(n=k)^oo a_n is also convergent.
If int_k^oo f(x) dx is divergent then the series sum_(n=k)^oo a_n is also divergent.
For the series sum_(n=1)^oo n*e^(-n/2) , we have a_n =n*e^(-n/2) then we may let the function:
f(x) =x*e^(-x/2) .
The graph of f(x) is:

As shown on the graph, f(x) is positive on the interval [1,oo) . Based on the behavior of the graph as x increases, the function eventually decreases. We can confirm this by applying First Derivative test. To determine the derivative of the function, we may apply the Product rule for differentiation: d/(dx) (u*v)= v* du+ u *dv .
Let: u =x then du = 1
v=e^(-x/2) then dv =- e^(-x/2)/2
Note: d/(dx)e^(-x/2) = e^(-x/2) * d/(dx) (-x/2)
=e^(-x/2) *(-1/2)
=- e^(-x/2)/2
Applying the Product rule for differentiation, we get:
f'(x) =e^(-x/2) * 1 + x *- e^(-x/2)/2
=e^(-x/2) - (xe^(-x/2))/2
= (e^(-x/2) (2-x))/2
Solve for critical values of x by applying f'(x) =0 .
(e^(-x/2) (2-x))/2 =0
(e^(-x/2) (2-x))=0
Apply zero-factor property:
(2-x)=0 then x=2
Using test point x=5 after x=2 , we get:
f'(5) = (e^(-5/2) (2-5))/2 ~~ -0.12313 .
When f'(x) lt0 , then the function is decreasing for the given integral.
Then f(x)=x*e^(-x/2) from the interval [2, oo) . Since the function is ultimately decreasing on the interval [1,oo) we may apply the integral test:
int_1^oo x*e^(-x/2) dx= lim_(n-gtoo) int_1^tx*e^(-x/2)dx
To determine the indefinite integral of int_1^t x*e^(-x/2)dx , we may apply u-substitution by letting: u =-x/2 or x=-2u then du = -1/2 dx or -2du =dx .
The integral becomes:
int x*e^(-x/2)dx=int (-2u)*e^u*(-2du)
= int 4ue^u du
= 4 int ue^udu
Apply the integration formula for exponential functions: int xe^xdx=(x-1)e^x+C.
4 int ue^udu=4 *(u-1)e^u
= 4ue^u -4e^u
Plug-in u =-x/2 on 4ue^u -4e^u , we get:
int_1^t x*e^(-x/2)dx =4(-x/2)e^(-x/2) -4e^(-x/2)|_1^t
=-2xe^(-x/2) -4e^(-x/2)|_1^t
Applying definite integral formula: F(x)|_a^b = F(b)-F(a).
-2e^(-x/2) -4e^(-x/2)|_1^t=[-2te^(-t/2) -4e^(-t/2)]-[-2*1e^(-1/2) -4e^(-1/2)]
=-2te^(-t/2) -4e^(-t/2)+2e^(-1/2) +4e^(-1/2)
=-2te^(-t/2) -4e^(-t/2)+6e^(-1/2)
Applying int_1^t x*e^(-x/2)dx =-2te^(-t/2) -4e^(-t/2)+6e^(-1/2) , we get:
lim_(n-gtoo) int_1^tx*e^(-x/2)dx =lim_(n-gtoo)[-2te^(-t/2) -4e^(-t/2)+6e^(-1/2)]
=lim_(n-gtoo)-2te^(-t/2) -lim_(n-gtoo)4e^(-t/2)+lim_(n-gtoo)6e^(-1/2)
=-2*ooe^(-oo) -4e^(-oo)+6e^(-1/2)
=0-0+6/e^(1/2)
=6/e^(1/2) or 6/sqrt(e)
The lim_(n-gtoo) int_1^tx*e^(-x/2)dx =6/sqrt(e) implies that the integral converges.
Conclusion:
The integral int_1^oo x*e^(-x/2)dx is convergent therefore the series sum_(n=1)^oo n*e^(-n/2) must also be convergent.