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In the most general form, for arbitrary ideals over rings, this is false. In the ring $\mathbb{Z}$ let $I_k$ be generated by $2^k$ and let $J$ be generated by $3$. Then $I_k+J=\mathbb{Z}$ for all $k$ and so $\cap_{k=1}^\infty(I_k+J)=\mathbb{Z}$. But $\bigcap_{k=1}^\infty I_k=\lbrace0\rbrace$ and so $J+\bigcap_{k=1}^\infty I_k=J\ne\mathbb{Z}$.