Convergence of the series n=3 to infinity 1/(logn)^log(logn)

Introduction:
In this problem, we need to determine whether the series given by the formula 1/(logn)^log(logn) converges or diverges. To do this, we will analyze the behavior of the series as n approaches infinity.

Analysis:
To determine the convergence of the series, we will use the comparison test. We will compare the given series with a known convergent or divergent series. Let's consider the series 1/n^2 as our comparison series.

Step 1: Taking the limit:
To compare the given series with the comparison series, we need to take the limit as n approaches infinity for both series. Let's start with the given series:

lim(n→∞) 1/(logn)^log(logn)

Step 2: Simplifying the expression:
To simplify the expression, we can use logarithmic properties. First, let's rewrite the expression as e^(log(1/(logn)^log(logn))). Using the properties of logarithms, we can simplify further:

e^(log(1/(logn)^log(logn))) = e^(log(1) - log(logn) * log(log(logn)))

Since log(1) is equal to 0, the expression simplifies to:

e^(-log(logn) * log(log(logn)))

Step 3: Taking the limit:
Now, let's take the limit as n approaches infinity:

lim(n→∞) e^(-log(logn) * log(log(logn)))

Step 4: Applying L'Hôpital's Rule:
To evaluate the limit, we can apply L'Hôpital's Rule. Differentiating both the numerator and denominator with respect to logn, we get:

lim(n→∞) (-1/(logn)) * log(log(logn))

Step 5: Simplifying the expression:
Further simplifying the expression, we have:

lim(n→∞) -log(log(logn))/logn

As n approaches infinity, log(log(logn)) and logn both grow without bound, but log(log(logn)) grows slower than logn. Therefore, the limit can be simplified to:

lim(n→∞) -1/logn

Step 6: Comparing with the comparison series:
Now, let's compare the limit we obtained with the comparison series 1/n^2. We have:

lim(n→∞) -1/logn < lim(n→∞)="" />

The series 1/n^2 is a convergent series (p-series with p = 2), which means that if the given series is smaller or equal to 1/n^2, it will also converge.

Conclusion:
Since the limit of the given series is smaller than the limit of the convergent series 1/n^2, we can conclude that the series n=3 to infinity 1/(logn)^log(logn) converges.