The image shows the following mathematical expression:

$\sum_{n=1}^{\infty} \frac{1}{n \log(n) \sqrt{\log^2 n - 1}}$

This is an infinite series where each term is given by:

$a_n = \frac{1}{n \log(n) \sqrt{\log^2 n - 1}}$

To analyze the convergence of this series, we would typically apply a convergence test such as the Comparison Test, Ratio Test, or Integral Test.

Convergence Analysis

1. Comparison Test: Comparing this series to a known convergent or divergent series.
2. Ratio Test: Examining the limit of the ratio of consecutive terms.
3. Integral Test: Converting the series to an improper integral and analyzing its convergence.

In this case, due to the complexity of the expression involving both logarithms and square roots, a good approach might be to approximate the behavior of the series for large $n$ and compare it with a simpler, known series.

For large $n$:

• $\log^2 n$ dominates over 1 in $\log^2 n - 1$, so the square root $\sqrt{\log^2 n - 1}$ is approximately $\log n$.
• Therefore, the general term $a_n$ can be approximated as $\frac{1}{n \log(n) \cdot \log n} = \frac{1}{n (\log n)^2}$.

This suggests a comparison with the series:

$\sum_{n=1}^{\infty} \frac{1}{n (\log n)^2}$

The series $\sum_{n=2}^{\infty} \frac{1}{n (\log n)^p}$ converges if and only if $p > 1$. Here, $p = 2$, so this comparison indicates that the original series converges.

Would you like a more detailed step-by-step breakdown of the convergence analysis, or do you have any specific questions?

Here are some related questions for further exploration:

1. How does the Integral Test apply to the given series, and what would it reveal?
2. Can we use the Ratio Test on this series, and what would the results indicate?
3. What are other series that can be compared using the Comparison Test?
4. How does changing the exponent in the logarithm affect the convergence of similar series?
5. What are some real-world applications of analyzing the convergence of series like this?

Tip: When dealing with logarithmic series, it’s often helpful to simplify by considering the dominant term in the logarithm for large $n$.