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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.

Show that the series ∑_{n ≥ 2} (-1)^n / [n(log n)^p] is absolutely convergent if p > 1.

In this example, learn how to show that the series ∑_{n ≥ 2} (-1)^n / [n(log n)^p] is absolutely convergent for p > 1 using the integral test. This problem involves series convergence, logarithmic functions, and advanced calculus techniques. Suitable for undergraduate-level calculus students.

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