`sum_(n=1)^oo n*e^(-n/2)` Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.

Expert Answers
marizi eNotes educator| Certified Educator

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.