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

4 Wadim Zudilinof degree n. Then(9)Q n (z −1 )F (z) ==n∑q n−µ z µ−n ·µ=0∞∑l=0z l−nn∑µ=0µ≤ln−1∑= r l z l−n +l=0∞∑( ) a + b + νz νbν=0q n−µ( a + b + l − µb)∞∑r l z l−n = P n (z −1 ) + R n (z).l=nHere the polynomial(10)n−1∑P n (x) = r l x n−l ∈ Z[x], where r l =l=0l∑µ=0has degree at most n, while the coefficients of the remainder∞∑R n (z) = r l z l−nl=n( )a + b + l − µq n−µ ,bare of the following form:n∑( )a + b + l − µr l = q n−µbµ=0n∑( )( )( )a + 2n − 1 − µ a + b + n a + b + l − µ= (−1) n−µ n − µµbµ=0= (−1) n (a + b + n)!n∑( ) n (a + 2n − 1 − µ)!(a + b + l − µ)!(−1) µ(a + n − 1)!n!b!µ (a + l − µ)!(a + b + n − µ)!µ=0= (−1) n (a + b + n)!(a + n − 1)!n!b!(a + 2n − 1)!(a + b + l)!n∑ (−n) µ (−a − l) µ (−a − b − n) µ×(a + l)!(a + b + n)! µ!(−a − 2n + 1)µ=0µ (−a − b − l) µ( n (a + 2n − 1)!(a + b + l)! −n, −a − l, −a − b − n= (−1) · 3F 2 (a + n − 1)!(a + l)!n!b! −a − 2n + 1, −a − b − l ∣). 1If we apply (4) with the choice A = −a−l, B = −a−b−n and C = −a−b−l,we obtainn (a + 2n − 1)!(a + b + l)!r l = (−1) ·(a + n − 1)!(a + l)!n!b!(−b) n (n − l) n(−a − b − l) n (a + n) n.