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

Answer 1

The series of sequence #u_n# is convergent.

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)# :
(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.

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Answer 2

To determine convergence or divergence for the summation of ( n \cdot e^{-n/2} ) using the integral test, we integrate the function ( f(x) = x \cdot e^{-x/2} ) from 1 to infinity. If the integral converges, then the series converges; if the integral diverges, then the series diverges.

The integral of ( f(x) ) can be evaluated as follows:

[ \int_{1}^{\infty} x \cdot e^{-x/2} , dx ]

Using integration by parts, with ( u = x ) and ( dv = e^{-x/2} ), we get:

[ du = dx ] [ v = -2e^{-x/2} ]

Applying the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

[ = -2xe^{-x/2} - \int (-2e^{-x/2} , dx) ]

[ = -2xe^{-x/2} + 4e^{-x/2} ]

Now, we evaluate this integral from 1 to infinity:

[ \lim_{t \to \infty} \left[ -2te^{-t/2} + 4e^{-t/2} \right] - \left[ -2e^{-1/2} + 4e^{-1/2} \right] ]

[ = 4e^{-1/2} ]

Since the value of the integral is finite, the series ( \sum_{n=1}^{\infty} n \cdot e^{-n/2} ) converges by the integral test.

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Answer 3

To determine convergence or divergence for the summation of ( n \cdot e^{-n/2} ) using the integral test, follow these steps:

  1. Integrate the function ( f(x) = x \cdot e^{-x/2} ) from 1 to infinity.
  2. Evaluate the definite integral.
  3. If the integral converges, then the series also converges. If the integral diverges, then the series also diverges.

The integral of ( f(x) ) from 1 to infinity is ( \int_1^\infty x \cdot e^{-x/2} , dx ). To solve this integral, you can use integration by parts.

Let ( u = x ) and ( dv = e^{-x/2} , dx ). Then, ( du = dx ) and ( v = -2e^{-x/2} ).

Apply the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

[ \int_1^\infty x \cdot e^{-x/2} , dx = -2x \cdot e^{-x/2} \Bigg|_1^\infty + 2 \int_1^\infty e^{-x/2} , dx ]

Evaluate the integral from 1 to infinity:

[ \lim_{b \to \infty} -2be^{-b/2} + 2\left(-2e^{-x/2}\right) \Bigg|_1^b ]

[ = \lim_{b \to \infty} -2be^{-b/2} + 4e^{-b/2} - (-2e^{-1/2} + 4e^{-1/2}) ]

[ = -2\lim_{b \to \infty} \left(be^{-b/2} - 2e^{-b/2}\right) + 4e^{-1/2} ]

[ = -2(0) + 4e^{-1/2} ]

[ = 4e^{-1/2} ]

Since the integral ( \int_1^\infty x \cdot e^{-x/2} , dx ) converges (equals ( 4e^{-1/2} )), the series ( \sum_{n=1}^\infty n \cdot e^{-n/2} ) also converges by the integral test.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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