Final answer:
To determine whether the series Σn=3 In(n)/4n converges or diverges, we can use the integral test. By evaluating the improper integral ∫ln(x)/(4x)dx from 3 to ∞, we find that it converges, indicating that the series also converges.
Step-by-step explanation:
To determine whether the series Σn=3 In(n)/4n converges or diverges, we can use the integral test. The integral test states that if f(x) is a positive, continuous, and decreasing function on the interval [n, ∞), then the series Σf(n) converges if and only if the corresponding improper integral ∫f(x)dx from n to ∞ converges. In this case, f(x) = ln(x)/(4x), which is positive, continuous, and decreasing for x ≥ 3. Therefore, we need to evaluate the improper integral ∫ln(x)/(4x)dx from 3 to ∞ to see if it converges.
We can evaluate the integral using integration by parts. Let u = ln(x) and dv = 1/(4x)dx. Then, du = 1/x dx and v = (1/4)ln(x). Applying the integration by parts formula, we have:
∫ln(x)/(4x)dx = (1/4)ln(x)ln(x) - ∫(1/4)(1/x)ln(x)dx
Simplifying further, we have:
∫ln(x)/(4x)dx = (1/4)ln(x)ln(x) - (1/4)∫ln(x)/(4x)dx
Re-arranging the equation, we get:
(5/4)∫ln(x)/(4x)dx = (1/4)ln(x)ln(x)
Now we can solve for the integral:
∫ln(x)/(4x)dx = (1/5)ln(x)ln(x)
Since the integral ∫ln(x)/(4x)dx from 3 to ∞ evaluates to a finite value, the series Σn=3 In(n)/4n converges.