To show that the series

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

converges absolutely, we can proceed as follows:

Step 1: Definition of Absolute Convergence

A series $\sum a_n$ converges absolutely if the series of its absolute values, $\sum |a_n|$, converges.

For the given series, the terms are $a_n = \frac{(-1)^{n-1}}{n^2}$, so the absolute values are:

$|a_n| = \frac{1}{n^2}.$

Step 2: Analyze the Series $\sum \frac{1}{n^2}$

The series $\sum \frac{1}{n^2}$ is a p-series with $p = 2$. Recall that a p-series $\sum \frac{1}{n^p}$ converges if $p > 1$. Since $p = 2 > 1$, the series $\sum \frac{1}{n^2}$ converges.

Step 3: Conclusion

Because the series $\sum \frac{1}{n^2}$ converges, the series $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}$ converges absolutely. Absolute convergence also implies regular convergence.

If you'd like a more detailed explanation or proof, let me know!

Related Questions:

1. What is the difference between absolute convergence and conditional convergence?
2. Can the alternating series test be used to show regular convergence of this series?
3. What is a p-series, and why does it converge for $p > 1$?
4. How would the convergence change if the denominator were $n^p$ with $0 < p \leq 1$?
5. Could we use a comparison test to prove the absolute convergence of this series?

Tip:

For series involving $n^p$, always first identify the type of series (p-series, geometric, etc.) to simplify the convergence analysis.