a[k]=t
a[k+1]=√(t+s)
f(t)=t+s-t^2=-(t-1/2)^2+s+1/4
2<s 2<tでf(t)<f(2)=2-s<0 t+s<t^2 a[k+1]<a[k]
0<s<2 0<t<2でf(t)>2-s>0 t+s>t^2 a[k+1]>a[k]