For k > 0 , let R k be the region in R 3 given by
R k = { ( x , y , z ) : ∣ x ∣ k + ∣ y ∣ k ≤ 1 AND ∣ y ∣ k + ∣ z ∣ k ≤ 1 AND ∣ z ∣ k + ∣ x ∣ k ≤ 1 } .
Let V k be the volume of R k . Let V ∞ = lim k → ∞ V k . There exists integers a , b , c such that
k → ∞ lim k a ( V ∞ − V k ) = b π c .
Submit a + b + c + V 1 + 1 0 V 1 / 2 + ( V 2 − 1 6 ) 2 .
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I got V 1 = 2 , V 1 / 2 = 3 / 1 0 , V 2 = 1 6 − 8 2 , V ∞ = 8 (these are pretty easy). I also got
V ∞ − V k ∼ 8 ⋅ 3 ( 1 − Γ ( k k + 2 ) Γ ( 1 + k 1 ) 2 ) ∼ 8 ⋅ 3 6 π 2 k − 2
but I am not 100% sure this is correct (I may have made a mistake).