Sigma with n=1 to n= positive infinity
(x^3)* (e^(-x^4))
Does it converge or diverge and use the integral test to show how. I got that it diverges, but this isn't correct. Can someone please explain? Thanks.
I assume the expression is
(n^3)* (e^(-n^4)) for n=1 to +&inf;
You can use the ratio test to show that it converges, namely, show that
an+1/an < 1 ∀ k≥1.
The integral test will result in a definite integral from 1 to +&inf; which is not an elementary function, but evaluates numerically to about 0.09. However, this still shows that the series is convergent.
Integral from 1 to infinity of
(x^3)* (e^(-x^4)) dx =
1/4 exp(-1)
To determine whether the series ∑(n=1 to ∞) (x^3)(e^(-x^4)) converges or diverges using the integral test, we need to check the convergence of the corresponding improper integral.
The integral test states that if f(x) is a positive, continuous, and decreasing function on the interval [1, ∞), then the series ∑(n=1 to ∞) f(n) and the integral ∫(1 to ∞) f(x) dx either both converge or both diverge.
Now, let's apply the integral test to the given series:
Step 1: Check if f(x) = x^3 * e^(-x^4) satisfies the conditions of the integral test.
a) Positivity: The function f(x) = x^3 * e^(-x^4) is always positive for any x > 0 since both terms in the product are positive.
b) Continuity: The function f(x) = x^3 * e^(-x^4) is a continuous function for all x > 0 as it is a product of two continuous functions.
c) Decreasing: To check if the function f(x) = x^3 * e^(-x^4) is decreasing, we can calculate its derivative:
f'(x) = 3x^2 * e^(-x^4) + (-4x^3) * x^3 * e^(-x^4)
= 3x^2 * e^(-x^4) - 4x^6 * e^(-x^4)
= e^(-x^4) (3x^2 - 4x^6)
To determine the sign of f'(x), we need to consider the sign of (3x^2 - 4x^6) since e^(-x^4) is positive for all x.
Setting (3x^2 - 4x^6) > 0 and solving for x, we find that x < 0 or 0 < x < 1, or x > 1. However, since we are only interested in the interval [1, ∞), we focus on x > 1.
For x > 1, (3x^2 - 4x^6) < 0, which means that f'(x) < 0 and the function f(x) = x^3 * e^(-x^4) is decreasing on the interval [1, ∞).
Therefore, f(x) = x^3 * e^(-x^4) satisfies all the conditions of the integral test.
Step 2: Evaluate the integral ∫(1 to ∞) f(x) dx to determine if it converges or diverges.
∫(1 to ∞) (x^3 * e^(-x^4)) dx
To evaluate this integral, we can make a substitution: u = -x^4, du = -4x^3 dx. Rearranging, we have dx = -du / (4x^3).
∫(1 to ∞) (x^3 * e^(-x^4)) dx = ∫(-1 to ∞) [(1/4)(-du / u) e^u]
Now, taking the limit of this integral as the upper bound approaches infinity:
lim (a→∞) ∫(-1 to a) [(1/4)(-du / u) e^u]
= lim (a→∞) [(1/4) * (e^u * ln|u|) |(-1 to a)]
= lim (a→∞) [(1/4) * [(e^a * ln|a|) - (e^(-1) * ln|-1|)]]
ln|-1| is undefined, but ln|u| approaches negative infinity as u approaches zero from the left. So, ln|-1| can be treated as negative infinity.
Taking the limit, we have:
= lim (a→∞) [(1/4) * [(e^a * ln|a|) - (e^(-1) * (-∞))]]
= lim (a→∞) [(1/4) * [(e^a * ln|a|) + ∞]]
Here, we have a combination of finite and infinite terms, resulting in an indeterminate form. Therefore, we cannot determine the exact value of the integral.
However, since the integral does not converge to a finite value, it diverges. This implies that the series ∑(n=1 to ∞) (x^3)(e^(-x^4)) also diverges.
Thus, the series diverges, which contradicts the claim that it converges. Therefore, your initial conclusion was correct.