>>195

1/(nCk) = (n+1)B(n+1-k,k+1) = (n+1)∫[0,1] t^k (1-t)^(n-k) dt (Β()はベータ関数)を代入
S[n] = (n+1)∫[0,1] Σ[k=0,n] t^k (1-t)^(n-k) dt
　　= 2(n+1)∫[0,1/2] ((1-t)^(n+1) - t^(n+1))/(1-2t) dt

t^(n+1)<(1/2)^(n+1), (1-2t)<(1-t)^2 (0<t<1/2) を代入
S[n] > (1+1/n)(2-(n+2)/2^n)

(1-t)^(n+1)<e(-(n+1)t), t^(n+1)>0, 1/(1-2t)<e^(3t) (0<t<1/4)
((1-t)^(n+1) - t^(n+1))/(1-2t) < 2(3/4)^(n+1) (1/4<t<1/2) を代入
S[n] < 2(n+1)/(n-2) + (n+1)(3/4)^(n+1)

(1) S[n]→2 (n→∞)

(2) p=1

(3) n=4,5,6を同時に満たすA,Bは存在しない