Analysis Converging or diverging

[sara_87]

Homework Statement

Discuss the (absolute) convergence or divergence of hthe series whose nth term is:

(ln(n))^(-ln(n))

Homework Equations

The Attempt at a Solution

I want to use the integral rule. First i show that it is decreasing by differentiating the function (and realising it will be negative for all x) then i integrate (ln(x))^(-ln(x)) and show that the integral is finite non zero.
But how do i integrate (ln(x))^(-ln(x)) ??
Thank you

 

[mighty2000]

for your post! The series whose nth term is (ln(n))^(-ln(n)) is known as the "logarithm series" and it is a well-known example of a series that diverges. In fact, it is a special case of the p-series with p = -ln(n), which is known to diverge for all p ≤ 1.

To show this using the integral test, we can first rewrite the nth term as (1/ln(n))^(ln(n)). Then, we can use the u-substitution method with u = ln(n) and du = (1/n) dx. This gives us the integral ∫(1/ln(n))^(ln(n)) dx = ∫(1/u)^u du.

Next, we can use the power rule to integrate this, giving us ∫(1/u)^u du = (1/u)^(u+1)/(u+1) + C.

Now, we can plug in u = ln(n) and evaluate the integral from n = 1 to infinity, giving us ∫(1/ln(n))^(ln(n)) dx = (1/ln(n))^(ln(n)+1)/(ln(n)+1) + C.

Since the base of the exponent is less than 1, (1/ln(n))^(ln(n)+1) will approach 0 as n approaches infinity. Therefore, the integral will approach 0 as well. This means that the integral is finite and nonzero, and by the integral test, the series diverges.

In conclusion, the series (ln(n))^(-ln(n)) diverges by the integral test, and therefore, it does not converge absolutely.