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

The series of sequence

We have the series of sequence

In order to use the integral test,

Let's study the sign of

(Again, the sign of

Thus,

So we can use the integral test.

The series of sequence

Of course, here

If we want to calculate the integral, we must firstly find the antiderivative of

The antiderivative of a function

We have

We know that the derivative of

So we can easily find the antiderivative of

So

Again, we can easily find the antiderivative of

Therefore :

(Note :

Therefore, the series of sequence

By signing up, you agree to our Terms of Service and Privacy Policy

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.

By signing up, you agree to our Terms of Service and Privacy Policy

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

- Integrate the function ( f(x) = x \cdot e^{-x/2} ) from 1 to infinity.
- Evaluate the definite integral.
- 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.

By signing up, you agree to our Terms of Service and Privacy Policy

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.

- How do you use the Integral test on the infinite series #sum_(n=1)^oo1/sqrt(n+4)# ?
- How do you use L'hospital's rule to find the limit #lim_(x->oo)ln(x)/sqrt(x)# ?
- How do you determine if the improper integral converges or diverges #int [ (x arctan x) / (1+x^2)^2 ] dx# from negative 0 to infinity?
- How do you use the direct Comparison test on the infinite series #sum_(n=2)^oon^3/(n^4-1)# ?
- How do you use L'hospital's rule to find the limit?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7