The image you've uploaded contains the mathematical expression: $\sum_{n=2}^{\infty} \frac{1}{(\ln n)^2}$

This expression is a summation (series) involving the natural logarithm of $n$. To analyze its convergence, we can consider the properties of the logarithm and compare this series to known convergent or divergent series.

Analyzing the Series

The series $\sum_{n=2}^{\infty} \frac{1}{(\ln n)^2}$ can be analyzed using the integral test for convergence. The integral test states that if $f(x)$ is a continuous, positive, decreasing function for $x \geq a$ and $\sum_{n=a}^\infty f(n)$ is our series, then the series converges if and only if the improper integral $\int_a^\infty f(x) \, dx$ converges.

Applying the Integral Test

Let's consider the function $f(x) = \frac{1}{(\ln x)^2}$, which is continuous, positive, and decreasing for $x > 1$. The corresponding integral is $\int_2^\infty \frac{1}{(\ln x)^2} \, dx$.

Calculating the Integral

To evaluate this integral, we use substitution. Let $u = \ln x$, then $du = \frac{1}{x} dx$ or $dx = x \, du = e^u \, du$. Substituting these into the integral gives us: $\int_{\ln 2}^\infty \frac{1}{u^2} e^u \, du$

This integral is a bit complex to evaluate directly, but we can observe its behavior:

1. As $u \rightarrow \infty$, $e^u$ grows exponentially while $\frac{1}{u^2}$ decreases polynomially.
2. This implies that the growth of $e^u$ dominates, and the integral likely diverges.

Conclusion

Therefore, the series $\sum_{n=2}^{\infty} \frac{1}{(\ln n)^2}$ diverges based on the integral test, where the integral $\int_2^\infty \frac{1}{(\ln x)^2} \, dx$ also diverges.

Would you like more details on this analysis or have any other questions?

Here are some further explorations you might consider:

1. How does this series compare to other series involving logarithmic terms?
2. What would change if the series started from a different value than 2?
3. How does the convergence behavior change if we alter the power of the logarithm, say to $(\ln n)^3$?
4. Can numerical methods be used to estimate the sum of the first few terms of this series?
5. What are the implications of the divergence of such a series in practical applications?

Tip: Remember, series involving logarithmic terms often require careful handling as logarithmic functions grow very slowly, influencing convergence and divergence in sometimes unintuitive ways.