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

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Answer 1 · 2017-08-17T00:20:20

$sum_(j=3)^5 = 124/429$

With so few term involved, the easiest approach is just write out the terms and compute the sum:

$sum_(j=3)^5 1/(j^2-3) = 1/(3^2-3) + 1/(4^2-3) + 1/(5^2-3)$
$" " = 1/(9-3) + 1/(16-3) + 1/(25-3)$
$" " = 1/(6) + 1/(13) + 1/(22)$
$" " = 124/429$

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