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1 2k + 1− 1 4k2 k# . When k= 1 then we have ρmax(1−ρ max)k =ρ max(1−ρ max)⩽1−ρ max. Therefore, for any value of k⩾1 , we have the following general upper bound: 1−τ 2 ⩽(1−ρ max)

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