Answer
$a_n = \frac{(-1)^{n+1}}{n^2} $
Work Step by Step
The sequence has terms that are reciprocals of squares of the positive integers, with alternating signs; $(-1)^{n+1}$ makes the signs alternate, and $n^2$ given squares in the denominator.
$a_1 = \frac{(-1)^{1+1}}{1^2}=1 $
$a_2 = \frac{(-1)^{2+1}}{2^2}=-\frac{1}{4} $
$a_3 = \frac{(-1)^{3+1}}{3^2}=\frac{1}{9} $
.....
$a_n = \frac{(-1)^{n+1}}{n^2} $
