Test for convergence or divergence. Indicate the test that was used and justify your answer.
Sigma (lower index n = 2; upper index infinity) [ln(n)]/n
To determine whether the series Sigma (lower index n = 2; upper index infinity) [ln(n)]/n converges or diverges, we can use the Integral Test.
First, let's state the Integral Test:
The Integral Test states that if f(x) is a positive, continuous, and decreasing function on the interval [1, ∞] and if a_n = f(n), then the series Sigma (lower index n = 1; upper index infinity) a_n converges if and only if the improper integral Int (lower limit 1; upper limit infinity) f(x) dx converges.
In our case, we have a_n = [ln(n)]/n. To apply the Integral Test, we need to consider the function f(x) = [ln(x)]/x.
Now, let's verify the conditions of the Integral Test:
1. Positivity: The function f(x) = [ln(x)]/x is positive for all x > 1 since both ln(x) and x are positive for x > 1.
2. Continuity: The function f(x) = [ln(x)]/x is continuous for all x > 1 since both ln(x) and x are continuous for x > 1.
3. Decreasing: To check if f(x) = [ln(x)]/x is decreasing, we can find its derivative and determine its sign. Taking the derivative, we get f'(x) = (1 - ln(x))/(x^2). Since f'(x) < 0 for x > e, where e is Euler's number, we can conclude that f(x) = [ln(x)]/x is decreasing for x > e.
Since the conditions of the Integral Test are satisfied, we can proceed with evaluating the improper integral Int (lower limit 1; upper limit infinity) [ln(x)]/x dx.
Integrating f(x) = [ln(x)]/x with respect to x, we have:
Int (lower limit 1; upper limit infinity) [ln(x)]/x dx = lim (upper limit t -> infinity) Int (lower limit 1; upper limit t) [ln(x)]/x dx = lim (upper limit t -> infinity) [ln(t)^2]/2
Applying the limit, we have:
lim (upper limit t -> infinity) [ln(t)^2]/2 = infinity
Since the improper integral diverges (equals infinity), the series Sigma (lower index n = 2; upper index infinity) [ln(n)]/n also diverges.
Therefore, the series diverges, as determined using the Integral Test, and there is no need to consider further convergence tests.