The sum \(\sum_{k=3}^{\infty} \frac{1}{k(k^4-5k^2+4)^2}\) is equal to \(\frac{m^2}{2n^2}\), where \(m\) and \(n\) are relatively prime positive integers. Find \(m+n\).

Let's think step by step and output the final answer within \boxed{}.
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