The given series is:

$\sum_{n \geq 2} (-1)^n \frac{1}{n (\log n)^p}$

We are tasked with showing that this series is absolutely convergent if $p > 1$.

Step-by-step solution:

1. Absolute Convergence: For absolute convergence, we need to check the convergence of the series of the absolute values of the terms:

   $\sum_{n \geq 2} \left| (-1)^n \frac{1}{n (\log n)^p} \right| = \sum_{n \geq 2} \frac{1}{n (\log n)^p}$

   This is now a non-alternating positive series.

2. Convergence Test: To determine the convergence of this series, we will use the integral test. The integral test states that if the function $f(n) = \frac{1}{n (\log n)^p}$ is positive, continuous, and decreasing for $n \geq 2$, then the convergence of the integral:

   $\int_2^\infty \frac{1}{x (\log x)^p} \, dx$

   is equivalent to the convergence of the series $\sum_{n \geq 2} \frac{1}{n (\log n)^p}$.

3. Integral Calculation: Let's calculate the improper integral:

   $\int_2^\infty \frac{1}{x (\log x)^p} \, dx$

   To solve this, we use the substitution $u = \log x$, which gives $du = \frac{1}{x} dx$. The limits of integration change accordingly:

   • When $x = 2$, $u = \log 2$,
   • When $x = \infty$, $u = \infty$.

   Thus, the integral becomes:

   $\int_{\log 2}^\infty \frac{1}{u^p} \, du$

   The integral $\int_{\log 2}^\infty \frac{1}{u^p} \, du$ converges if and only if $p > 1$, because for $p = 1$, it would diverge as a logarithmic integral, and for $p < 1$, the exponent is too small for convergence.

4. Conclusion: The series $\sum_{n \geq 2} \frac{1}{n (\log n)^p}$ converges if $p > 1$ based on the integral test. Hence, the original series is absolutely convergent for $p > 1$.

Would you like further details on any part of the explanation?

Here are some related questions:

1. How does the integral test work for series convergence?
2. Why is the substitution $u = \log x$ helpful in the integral calculation?
3. What is the difference between absolute convergence and conditional convergence?
4. What happens to the series if $p = 1$?
5. How would you handle this series if it didn't alternate (i.e., without the $(-1)^n$ factor)?

Tip: The integral test is a powerful tool for series with logarithmic or power terms. Always check the conditions where the function behaves continuously and decreases to apply it.