Answer
Converges Absolutely
Work Step by Step
First use alternating series test to determine convergence
$\Sigma^{\infty}_{n=1} \frac{(-1)^{n+1}}{n^2} $
i.) $ a_{n+1} = \frac{1}{(n+1)^2} \leq \frac{1}{n^2} = a_n$
ii.) $\lim\limits_{n \to \infty} \frac{1}{n^2} =0 $
Series Converges by Alternating Series Test
Now test for absolute or conditional convergence
$|\frac{(-1)^{n+1}}{n^2}| \leq \frac{1}{n^2} $
$\Sigma^{\infty}_{n=1} \frac{1}{n^2}$, Converges by p-series, so by the Direct Comparison Test the series Converges Absolutely
