Final answer:
The convergence of the series ∑ₙ=2[infinity] 1/n ln(3n) can be tested using the Integral Test on the function f(x)=1/x ln(3x), which is positive and decreasing on the interval [2, ∞). The integral ∫ 1/(x ln(3x)) dx from 2 to ∞ will determine the series' convergence; if the integral converges, so does the series.
Step-by-step explanation:
To determine whether the series ∑ₙ=2[infinity] 1/n ln(3n) converges or diverges, we can compare it to a known convergent or divergent series. Since the series involves a logarithmic function, let's consider the integral test for convergence. We have the function f(x)=1/x ln(3x) that we will integrate to determine the convergence of the series.
First, let's establish the interval on which the function f is positive:
f(x) = 1/x ln(3x) > 0 for all x > 1.
This interval is due to the fact that both 1/x and ln(3x) are positive for x > 1. Now, let's consider the Integral Test. If f is continuous, positive, and decreasing on [2, ∞), and if the integral of f from 2 to ∞ is finite, then the series converges. Otherwise, it diverges. The Integral Test can be applied since f decreases on the interval [2, ∞) as both the functions 1/x and ln(x) are decreasing for x > 1.
Let us compute the integral:
∫ f(x) dx = ∫ ⁱ/(x ln(3x)) dx from x=2 to ∞. If this improper integral converges, then the series does too. Otherwise, it diverges.
To solve the integral, we would typically use integration techniques such as substitution or integration by parts. However, without going into the specifics of the integration, we can reference the integral test for the convergence of a series, stating that if such an integral converges, the corresponding series converges as well.