Question

Use the integral test to determine if the series ( sum_{n=1}} n e{-n²} ) converges or diverges. Be sure to check any necessary conditions. 3. (3 points) Use the direct complete?

Answer 1

To assess the convergence of the series Σn=1∞ (n e^{-n^2}), the integral test is used. The function f(n) = n e^{-n^2} meets the criteria for the integral test, being continuous, positive, and decreasing for n ≥ 1. Evaluation of the corresponding improper integral confirms convergence, indicating the series also converges.

Step-by-step explanation:

To determine whether the series ∑_{n=1}^{∞} n e^{-n^2} converges, we can use the integral test. The integral test states that if f(n) is continuous, positive, and decreasing for n ≥ 1 and if the integral of f from 1 to infinity is finite, then the corresponding series converges.

First, we must ensure that the function f(n) = n e^{-n^2} meets these criteria:

• Continuity: The function is continuous for all n ≥ 1.
• Positivity: n e^{-n^2} > 0 for all n ≥ 1.
• Decreasing: The derivative f'(n) = e^{-n^2} (1 - 2n^2) is negative for n > 1/√2, showing that f(n) is decreasing for n ≥ 1.

Now, we evaluate the improper integral:

∫_1^{∞} n e^{-n^2} dn

We can use a substitution u = n^2, which gives us du = 2n dn. The integral becomes:

∫_{1}^{∞} 0.5 e^{-u} du

The evaluation of this integral shows that it converges to a finite value. Therefore, according to the integral test, the series ∑_{n=1}^{∞} n e^{-n^2} also converges.