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

6 Wadim Zudilinand from (8), (11) we conclude that the only negative power of z originateswith the last summand:− Q n+1 (z −1 )R n (z)( )a + 2n + 1= (−1) n z −n−1( 1 + O(z) ) ( ) a + b + n·z n( 1 + O(z) )a + nb − n( )( ) a + 2n + 1 a + b + n 1= (−1) n + O(1) as z → 0. □a + n b − n z4. Arithmetic constituentsWe begin this section by noting that, for any prime p > √ N,⌊ ⌋( )N Nord p N! = and ord p N = λ ,ppwhereλ(x) = 1 − {x} − {−x} = 1 + ⌊x⌋ + ⌊−x⌋ ={1 if x ∈ Z,0 if x /∈ Z.For primes p > √ a + b + n, let( {e p = min − − a + n } {+ − a + n + µ } { } µ(13)+µ∈Z pp p{ } { } { })a + b + n a + b + µ n − µ−++ppp(⌊= min − a + n ⌋ ⌊− − a + n + µ ⌋ ⌊ ⌋ µ−µ∈Z pp p⌊ ⌋ ⌊ ⌋ ⌊ ⌋)a + b + n a + b + µ n − µ+−−ppp( )( )≤ min ord a + n a + n + µ a + b + np0≤µ≤n a + n + µ µ n − µ( )( )a + n − 1 + µ a + b + n= min ord p0≤µ≤n µ n − µand(14)( { } { } { }a + n + µ a + n µe ′ p = min −+ +µ∈Z pp p{ } { } { })a + b + n a + b + µ n − µ−++ppp