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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.

Evaluate the convergence or divergence of the series ∑ from n=2 to ∞ of 1/(ln n)^2.

Analyze the convergence of the series ∑ from n=2 to ∞ of 1/(ln n)^2 using the integral test. The series involves logarithmic functions, which can make determining convergence complex. Learn the key steps, substitution methods, and the final result showing the divergence of the series.

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