28 Jun
2006
28 Jun
'06
2:31 p.m.
It does. sum_{k=1}^n C_k/4^k = A008549(n)/4^n = (4^n - (2n+1)C_n)/4^n ~ 1 - 1/sqrt(pi*n). Franklin T. Adams-Watters -----Original Message----- From: franktaw@netscape.net Does sum_{k>0} C_k/4^k = 1? It definitely converges (C_k ~ 4^k/(sqrt(pi)*k^3/2)), and the limit is close to 1.