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A new lower bound for (3/2)k - Wadim Zudilin

A new lower bound for (3/2)k - Wadim Zudilin

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12 <strong>Wadim</strong> <strong>Zudilin</strong>6. Related resultsThe above construction allows us to prove similar results <strong>for</strong> the sequences‖(4/3) k ‖ and ‖(5/4) k ‖ using the representations( 4 2(b+1) (= 23) b+1 1 + 1 ) −(b+1)(28)8(≡ (−1) a 2 b−3a+1 F a, b; − 1 )(mod Z),8and(29)where 3a ≤ b,( 5 7b+3 (= 2 · 54) b 1 + 3 ) −(2b+1)125(≡ (−1) a 2 · 3 a · 5 b−3a F a, 2b; − 3 )(mod Z),125where 3a ≤ b.Namely, taking a = 5m, b = 15m, n = 6m(+1) in case (28) and a = 3m,b = 9m, n = 7m(+1) in case (29) and repeating the arguments of Section 5,we arrive atTheorem 2. The following estimates are valid:∥ 4 k ∥ ∥∥∥∥(> 0.49143) k = 3 −k·0.64672207... <strong>for</strong> k ≥ K 1 ,∥ 5 k ∥ ∥∥∥∥(> 0.51524) k = 4 −k·0.47839775... <strong>for</strong> k ≥ K 2 ,where K 1 , K 2 are certain effective constants.The general case of the sequence ‖(1 + 1/N) k ‖ <strong>for</strong> an integer N ≥ 5 maybe treated as in [4] and [2] by using the representation( ) N + 1 b+1 (= 1 − 1 ) −(b+1) ( )1≡ F 0, b; (mod Z).NN + 1N + 1The best result in this direction belongs to M. Bennett [2]: ‖(1 + 1/N) k ‖ >3 −k <strong>for</strong> 4 ≤ N ≤ k3 k .Remark 3. As mentioned by the anonymous referee, our result <strong>for</strong> ‖(4/3) k ‖is of special interest. It completes Bennett’s result [3] on the order of theadditive basis {1, N k , (N + 1) k , (N + 2) k , . . . } <strong>for</strong> N = 3 (case N = 2 correspondsto the classical Waring’s problem); to solve this problem one needsthe <strong>bound</strong> ‖(4/3) k ‖ > (4/9) k <strong>for</strong> k ≥ 6. Thus we remain verification of the<strong>bound</strong> in the range 6 ≤ k ≤ K 1 .

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