We change the order of summation: \[ \sum_{k=2}^\infty \sum_{j=2}^{2008} \frac{1}{j^k} = \sum_{j=2}^{2008} \sum_{k=2}^\infty \frac{1}{j^k} = \sum_{j=2}^{2008} \frac{1}{j^2(1 - \frac{1}{j})} = \sum_{j=2}^{2008} \frac{1}{j(j-1)} = \sum_{j=2}^{2008} \displaystyle \left( \frac 1 {j-1} - \frac 1 j \displaystyle \right) = 1 - \frac 1 {2008} = \boxed{\frac{2007}{2008}}.\]