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

A new lower bound for ‖(3/2) k ‖ 5The assumed condition n ≤ b guarantees that the coefficients r l do notvanish identically (otherwise (−b) n = 0). Moreover, (n − l) n = 0 for lranging over the set n ≤ l ≤ 2n − 1, therefore r l = 0 for those l, whileFinally,(11)(a + 2n − 1)!(a + b + l)!r l =(a + n − 1)!(a + l)!n!b!b!/(b − n)! · (l − n)!/(l − 2n)!×(a + b + l)!/(a + b + l − n)! · (a + 2n − 1)!/(a + n − 1)!(a + b + l − n)!(l − n)!= for l ≥ 2n.n!(b − n)!(a + l)!(l − 2n)!R n (z) =∞∑∑∞r l z l−n = z n r ν+2n z νl=2n= z n 1n!(b − n)!= z n ( a + b + nb − n∞∑ν=0)ν=0(a + b + n + ν)!(n + ν)!z νν!(a + 2n + ν)!( a + b + n + 1, n + 1· 2F 1 a + 2n + 1 ∣). zUsing the integral (5) for the polynomial (8) and remainder (11) we arriveatLemma 1. The following representations are valid:andQ n (z −1 ) =R n (z) =(a + b + n)!(a + n − 1)!n!(b − n)!∫(a + b + n)!1(a + n − 1)!n!(b − n)! zn0∫ 10t a+n−1 (1 − t) b−n (1 − z −1 t) n dtt n (1 − t) a+n−1 (1 − zt) −(a+b+n+1) dt.We will also require linear independence of a pair of neighbouring Padéapproximants, which is the subject ofLemma 2. We have(12) Q n+1 (x)P n (x) − Q n (x)P n+1 (x) = (−1) n ( a + 2n + 1a + n)( a + b + nb − n)x.Proof. Clearly, the left-hand side in (12) is a polynomial; its constant termis 0 since P n (0) = P n+1 (0) = 0 by (10). On the other hand,Q n+1 (z −1 )P n (z −1 ) − Q n (z −1 )P n+1 (z −1 )= Q n+1 (z −1 ) ( Q n (z −1 )F (z) − R n (z) )− Q n (z −1 ) ( Q n+1 (z −1 )F (z) − R n+1 (z) )= Q n (z −1 )R n+1 (z) − Q n+1 (z −1 )R n (z),