Answer
$R=1$ ; interval of convergence is $[-1,1]$
Work Step by Step
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 $x=1$, the series converges by the alternating series test. At $x=-1$, the given series is a p-series and hence converges.
Hence, $R=1$ ; interval of convergence is $[-1,1]$
