How do you determine convergence or divergence for the summation of `n*e^(-n/2)` using the integral test how do you answer?

We have the series of sequence `u_n = n e^(-n/2)=n/sqrt(e^n)`.

In order to use the integral test, `f(n) = u_n` must be positive and decreasing on `[p;+oo[`.

Let's study the sign of `f(n)` :

(Note : `e^(-n/2)` is always positive and it has the value of zero only if `n->+oo`; therefore, the sign of `f(n)` depends only on the numerator : `n`.)

Thus, `f(n)` is positive on `]0; +oo[`.

Let's study the slope (the growth) of `f(n)`, which comes down to study the sign of `f'(n)`.

The derivative of a function `f(x) = g(x)*h(x)` is :

`f'(x) = g'(x)h(x) + g(x)h'(x)`.

Therefore :

`f(n) = n * e^(-n/2) = g(n)*h(n)`

`f'(n) = (n)' * e^(-n/2) + n * (e^(-n/2))'`

`f'(n) = e^(-n/2) + n * (-n/2)' * (e^(-n/2))`

`f'(n) = e^(-n/2) + n * (-1/2) * (e^(-n/2))`

`f'(n) = e^(-n/2)*(1-n/2)`

(Again, the sign of `f'(n)` depends only on the numerator : `1-n/2`).

Thus, `f(n)` is positive and decreasing on `[+2;+oo[`.

So we can use the integral test.

The series of sequence `u_n` converges if `int_p^(+oo) f(x) dx` exists.

Of course, here `f(x) = e^(-x/2) x` and `p = 2`.

If we want to calculate the integral, we must firstly find the antiderivative of `f(x)`.

The antiderivative of a function `f(x) = g'(x)h(x)` is :

`F(x) = g(x)h(x) - intg(x)h'(x)dx`

We have `f(x) = e^(-x/2) * x = g'(x) * h(x)`

We know that the derivative of `e^(-x/2)` is `-1/2e^(-x/2)`.

So we can easily find the antiderivative of `g'(x) = e^(-x/2)` :

`g(x) = -2*e^(-x/2)`.

So `F(x) = (-2*e^(-x/2)*x) - (int-2*e^(-x/2)*1)`

Again, we can easily find the antiderivative of `-2*e^(-x/2)`, which is `4*e^(-x/2)`.

`F(x) = (-2*e^(-x/2)*x) - (4*e^(-x/2)) = -2e^(-x/2)*(x+2) = -(2(x+2))/sqrt(e^x)`.

Therefore :

`int_p^(+oo) f(x) dx = [F(x)]_2^(+oo) = ''F(+oo)'' - F(2) = 0 +(2(2+2))/sqrt(e^2) = 8/e`.

(Note : `''F(+oo)'' = 0` since `sqrt(e^x)` increases faster to `+oo` than `x`).

Therefore, the series of sequence `u_n` is convergent.