Frobenius & Special Functions · Superchallenge problem

Let $j_1 < j_2 < \cdots$ be the positive zeros of $J_0(x) = \displaystyle\sum_{k\ge0}\frac{(-1)^k}{(k!)^2}\left(\frac x2\right)^{2k}$, and assume $J_0(x) = \displaystyle\prod_{k\ge1}\left(1 - \frac{x^2}{j_k^2}\right)$. Compute $\displaystyle\sum_{k\ge1}\frac{1}{j_k^4}$. Express your answer as a common fraction.