# How do you find the sum of Sigma 1/(j^2-3) where j is [3,5]?

Aug 17, 2017

${\sum}_{j = 3}^{5} = \frac{124}{429}$
${\sum}_{j = 3}^{5} \frac{1}{{j}^{2} - 3} = \frac{1}{{3}^{2} - 3} + \frac{1}{{4}^{2} - 3} + \frac{1}{{5}^{2} - 3}$
$\text{ } = \frac{1}{9 - 3} + \frac{1}{16 - 3} + \frac{1}{25 - 3}$
$\text{ } = \frac{1}{6} + \frac{1}{13} + \frac{1}{22}$
$\text{ } = \frac{124}{429}$