## Calculus 8th Edition

$R=1$ ; interval of convergence is $[-1,1]$
Let $a_{n}=(-1)^{n}\frac {x^{n}}{n^{2}}$, then $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{(-1)^{n+1}\frac {x^{n+1}}{(n+1)^{2}}}{(-1)^{n}\frac {x^{n}}{n^{2}}}|$ $=|x|\lt 1$ At $x=1$ , alternating series, thus converges and $x=-1$ the given series is a p-series , thus convergent. Hence, $R=1$ ; interval of convergence is $[-1,1]$