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

1 Answer
Jun 30, 2015

Answer:

The series of sequence #u_n# is convergent.

Explanation:

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)# :
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(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)#

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(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.