To analyze the nature of the given series:

$\sum_{n \geq 2} \frac{1}{\ln n}$

We need to determine whether the series converges or diverges. Let us proceed step-by-step:

Step 1: General Term Analysis

The general term of the series is: $a_n = \frac{1}{\ln n}$ As $n \to \infty$, $\ln n \to \infty$, so $a_n \to 0$. However, this is only a necessary condition for convergence, not a sufficient one.

Step 2: Comparison Test

For comparison, note that $\ln n$ grows slower than any polynomial of $n$. Hence, $\frac{1}{\ln n}$ grows slower than $\frac{1}{n}$ but decays very slowly. We compare it with the harmonic series, $\sum \frac{1}{n}$, which diverges.

To refine this, we consider an integral test.

Step 3: Integral Test

The series can be approximated by the integral: $\int_{2}^\infty \frac{1}{\ln x} \, dx$

Using the substitution $u = \ln x$, so $du = \frac{dx}{x}$, the integral becomes: $\int \frac{1}{\ln x} dx = \int \frac{1}{u} du = \ln |\ln x| + C$

Now, evaluate the improper integral: $\int_{2}^\infty \frac{1}{\ln x} \, dx = \ln (\ln x) \Big|_2^\infty$

As $x \to \infty$, $\ln (\ln x) \to \infty$. Hence, the integral diverges.

Step 4: Conclusion

By the integral test, since the corresponding integral diverges, the series: $\sum_{n \geq 2} \frac{1}{\ln n}$ diverges.

Let me know if you would like more detailed steps or additional clarifications!

Related Questions:

1. Why does the integral test work for this series?
2. How does $\frac{1}{\ln n}$ compare with $\frac{1}{n^p}$ for $p > 0$?
3. Can the divergence of this series be intuitively explained by the growth of $\ln n$?
4. How does this series relate to the harmonic series in terms of divergence speed?
5. Is there a slower-growing divergent series than $\frac{1}{\ln n}$?

Tip: When applying the integral test, always check whether the function is positive, continuous, and decreasing!