use the integral test to determine whether the series is convergent or divergent: n^2/e^n
from n = 3 to infinity
so plug and chug. Using
∫[3,∞] x^2/e^x dx
do this using integration by parts (twice), since it is just
∫[3,∞] x^2 e^(-x) dx = (x^2+2x+2) e^(-x)
so, does that integral converge or diverge?
To determine the convergence or divergence of the series ∑(n^2/e^n), we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function on the interval [1, ∞) such that f(n) = a_n, then the series ∑a_n converges if and only if the improper integral ∫(1 to ∞)f(x) dx converges.
In this case, we have a_n = n^2/e^n. Let's write the function f(x) = x^2/e^x.
Now, we need to check if f(x) satisfies the conditions of the integral test:
1. Positive: f(x) = x^2/e^x is positive for all x ≥ 1.
2. Continuous: Both x^2 and e^x are continuous functions, so their ratio x^2/e^x is also continuous on the interval [1, ∞).
3. Decreasing: Take the derivative of f(x) to find f'(x) = (2x - x^2)/e^x. To determine where f(x) is decreasing, we need to find where f'(x) < 0:
(2x - x^2)/e^x < 0
2x - x^2 < 0
x(2 - x) < 0
x(2 - x) = 0 at x = 0 and x = 2, so we have two critical points.
For x < 0, x(2 - x) is positive.
For 0 < x < 2, x(2 - x) is negative.
For x > 2, x(2 - x) is positive.
Therefore, f(x) = x^2/e^x is decreasing on the interval [1, ∞).
Since f(x) satisfies all the conditions of the integral test, we can evaluate the improper integral to determine the convergence or divergence of the series.
∫(1 to ∞)x^2/e^x dx = ∫(1 to ∞)x^2e^(-x) dx
To evaluate this integral, you can use integration by parts.
Let u = x^2 and dv = e^(-x) dx. Then, du = 2x dx and v = -e^(-x).
Using the integration by parts formula, we have:
∫(1 to ∞)x^2e^(-x) dx = -x^2e^(-x) - 2∫(1 to ∞)xe^(-x) dx
Next, we can apply integration by parts again to evaluate ∫xe^(-x) dx.
Let u = x and dv = e^(-x) dx. Then, du = dx and v = -e^(-x).
Using the integration by parts formula again, we have:
∫xe^(-x) dx = -xe^(-x) - ∫(-e^(-x)) dx
= -xe^(-x) + e^(-x)
Now, we can substitute these results back into the previous equation:
-∫(1 to ∞)x^2e^(-x) dx = -x^2e^(-x) - 2(-xe^(-x) + e^(-x))
= -x^2e^(-x) + 2xe^(-x) - 2e^(-x)
To evaluate the improper integral, we calculate the limits as x approaches ∞ and 1:
lim (x→∞) -x^2e^(-x) + 2xe^(-x) - 2e^(-x)
lim (x→∞) -∞ + 2(0) - 0 = -∞
lim (x→1) -x^2e^(-x) + 2xe^(-x) - 2e^(-x) = 1 - 2 + 2 = 1
As the limit as x approaches ∞ of the improper integral is -∞ and the limit as x approaches 1 is 1, the integral ∫(1 to ∞)x^2/e^x dx is divergent.
Therefore, by the integral test, the series ∑(n^2/e^n) is divergent.
To use the integral test, we need to check the convergence of the series by comparing it to an improper integral. For the series given, n^2/e^n, let's proceed with the integral test.
1. First, let's set up the integral: ∫ (n^2/e^n) dn.
2. To evaluate this integral, we use integration by parts. Integrating n^2 and differentiating e^n, we have:
∫ (n^2/e^n) dn = -n^2e^(-n) - 2∫ (ne^(-n)) dn.
3. Now, let's evaluate the second integral above using integration by parts again. Integrating ne^(-n) and differentiating n, we have:
∫ (ne^(-n)) dn = -ne^(-n) - ∫ (e^(-n)) dn.
4. The integral of e^(-n) is simply -e^(-n). Substituting this into the above expression, we get:
-ne^(-n) - ∫ (e^(-n)) dn = -ne^(-n) + e^(-n) + C,
where C is the constant of integration.
5. Returning to the original integral, we have:
∫ (n^2/e^n) dn = -n^2e^(-n) - 2(-ne^(-n) + e^(-n) + C).
Simplifying, we have:
∫ (n^2/e^n) dn = n^2e^(-n) + 2ne^(-n) - 2e^(-n) - 2C.
6. Now, we need to evaluate this integral from 1 to infinity to determine convergence.
∫∞ 1 (n^2/e^n) dn = lim (n -> ∞) [-n^2e^(-n) - 2ne^(-n) + 2e^(-n) - 2C] - [-1^2e^(-1) - 2e^(-1) + 2e^(-1) - 2C].
Simplifying further, we have:
= lim (n -> ∞) [-n^2e^(-n) - 2ne^(-n) + 2e^(-n)] - [-(e^(-1) + 2e^(-1) - 2C)].
7. Now, we need to determine the limit of the above expression as n approaches infinity.
Taking the limit, the terms involving e^(-n) will approach 0 as n goes to infinity. Therefore, the limit simplifies to:
= -(-(e^(-1) + 2e^(-1) - 2C)).
= (e^(-1) + 2e^(-1) - 2C).
8. So, the improper integral converges to (e^(-1) + 2e^(-1) - 2C) as n approaches infinity.
9. Now, we can conclude based on the integral test: If the integral of the series converges, then the series also converges. In this case, since the improper integral converges, the series n^2/e^n also converges.
Therefore, the series n^2/e^n is convergent.