Introduction:
We need to determine whether the series ∑(n=1 to ∞) 1/(log log n)^(log n) is convergent or divergent.

Test for convergence:
To determine the convergence or divergence of the given series, we can use the integral test.

Integral test:
The integral test states that if a function f(x) is continuous, positive, and decreasing on the interval [a, ∞), then the series ∑(n=1 to ∞) f(n) and the integral ∫(a to ∞) f(x) dx either both converge or both diverge.

Applying the integral test:
Let's consider the function f(x) = 1/(log log x)^(log x). We will test the convergence of the series by evaluating the integral of this function.

Step 1: Determine the interval:
The given series starts from n = 1, so we need to find the integral from 1 to ∞.

Step 2: Evaluate the integral:
∫(1 to ∞) 1/(log log x)^(log x) dx

Step 3: Apply substitution:
Let u = log x
Then, du = (1/x) dx
dx = x du

∫(1 to ∞) 1/(log log x)^(log x) dx
= ∫(0 to ∞) 1/(log log e^u)^u * x du
= ∫(0 to ∞) 1/(u log log e)^u * e^u du
= ∫(0 to ∞) 1/(u^u) du

Step 4: Evaluate the integral:
We can solve this integral using a comparison test.

Let's consider the function g(u) = 1/u^u.

Taking the derivative of g(u):
g'(u) = (1/u^u) * (d/dx(u^u))
= (1/u^u) * (u^u * (ln u + 1))
= (ln u + 1)/u^u

We can observe that g'(u) > 0 for all u > 1. Therefore, g(u) is a decreasing function for u > 1.

Comparison test:
Let's compare g(u) = 1/u^u with the function h(u) = 1/u^(u+1).

Taking the ratio of g(u) to h(u):
[g(u)/h(u)] = [1/u^u] / [1/u^(u+1)]
= u^(u+1) / u^u
= u

Taking the limit as u approaches ∞:
lim(u→∞) u = ∞

Since the limit is a finite positive value, we can conclude that g(u) and h(u) have the same behavior.

Step 5: Conclude the result:
We have shown that the integral ∫(1 to ∞) 1/(log log x)^(log x) dx is divergent.

Therefore, by the integral test, the series ∑(n=1 to ∞) 1/(log log n)^(log