Answer
$\displaystyle \sum_{j=3}^{5}\frac{1}{j^2-3}=\frac{124}{429}$
Work Step by Step
$\displaystyle \sum_{j=3}^{5}\frac{1}{j^2-3}=\frac{1}{3^2-3}+\frac{1}{4^2-3}+\frac{1}{5^2-3}=\frac{1}{6}+\frac{1}{13}+\frac{1}{22}=\frac{143+66+39}{858}=\frac{248}{858}=\frac{124}{429}$
