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.