Let's analyze the given series: \[ \sum_{n=0}^{\infty} (-1)^n \frac{x^n}{n^2} \] We are asked whether it **converges absolutely**, **converges conditionally**, or **diverges**. --- ### **Step 1: Test for Absolute Convergence** Absolute convergence means we consider: \[ \sum_{n=0}^{\infty} \left|(-1)^n \frac{x^n}{n^2}\right| = \sum_{n=0}^{\infty} \frac{|x|^n}{n^2} \] This is a power series, and for a general \( |x| < 1 \), the terms \(\frac{|x|^n}{n^2}\) resemble a p-series in the denominator times a geometric series in the numerator. #### **Check for convergence:** - For \( |x| < 1 \): - As \( n \to \infty \), \( |x|^n \to 0 \) rapidly. - \( n^2 \) grows faster than any polynomial in \( n \). - In fact, for any fixed \( |x| < 1 \), this converges by the **comparison test** to a convergent p-series, since \( \sum \frac{1}{n^2} \) converges, and \( |x|^n \) decays faster than any polynomial grows. - For \( |x| = 1 \): - The series becomes \( \sum \frac{1}{n^2} \), which converges (p-series with \( p = 2 > 1 \)). - Thus, the series converges absolutely for \( |x| \leq 1 \). - For \( |x| > 1 \): - \( |x|^n \) grows faster than \( n^2 \), so the terms do **not** approach zero, and the series **diverges**. --- ### **Step 2: Conclusion** - The given series **converges absolutely** for \( |x| \leq 1 \). - It **diverges** for \( |x| > 1 \). --- ### **Final Answer** **The series** \[ \sum_{n=0}^{\infty} (-1)^n \frac{x^n}{n^2} \] **converges absolutely for \( |x| \leq 1 \), and diverges for \( |x| > 1 \).** --- If you want to specify for a particular value of \( x \), let me know! For example, for \( x = 1 \), the series converges absolutely since it's a p-series with \( p=2 \).