The given series is:

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

Task:

We are asked to investigate whether this series converges.

Step-by-step analysis:

1. Form of the Series: The series alternates in sign due to $(-1)^n$. This suggests that we may apply the Alternating Series Test (also known as the Leibniz criterion) to check for convergence.

2. Alternating Series Test (Leibniz Criterion): The alternating series test requires two conditions to be met:

   • $a_n = \frac{1}{n^2 \sqrt{n^2 - 1}}$ must be decreasing.
   • $\lim_{n \to \infty} a_n = 0$.

Condition 1: $a_n$ is decreasing

We need to check if the sequence $a_n = \frac{1}{n^2 \sqrt{n^2 - 1}}$ is decreasing. Intuitively, as $n$ increases, both $n^2$ and $\sqrt{n^2 - 1}$ grow larger, making $a_n$ smaller. Therefore, $a_n$ should be decreasing for large $n$.

Condition 2: $\lim_{n \to \infty} a_n = 0$

Let's compute the limit of $a_n$ as $n \to \infty$:

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{n^2 \sqrt{n^2 - 1}}.$

For large $n$, we can approximate $\sqrt{n^2 - 1} \approx n$, so:

$a_n \approx \frac{1}{n^2 \cdot n} = \frac{1}{n^3}.$

Since $\lim_{n \to \infty} \frac{1}{n^3} = 0$, we have:

$\lim_{n \to \infty} a_n = 0.$

Thus, the second condition is satisfied.

Conclusion:

Since both conditions of the Alternating Series Test are satisfied (i.e., $a_n$ is decreasing and $\lim_{n \to \infty} a_n = 0$), we conclude that the series converges.

Would you like to go into further details, or do you have any questions?

Here are 5 related questions for further exploration:

1. How does the Alternating Series Test compare to other convergence tests?
2. Can we determine the sum of this convergent series?
3. What happens if we remove the alternating sign from the series?
4. How does the behavior of $a_n$ affect the speed of convergence?
5. Can we apply the ratio test or root test to this series for confirmation?

Tip: In general, the alternating series test is a reliable tool for establishing convergence when the series changes sign regularly.