Question

一道数列题 设0<a1<1/2,a(n+1)=f(an),n∈N*,证明an<1/(n+1)

f(x)=x-3/2 x^2

Answer 1

f(x)=x-3/2*x^2=-3/2(x^2-2/3x)=-3/2(x-1/3)^2+1/6<=1/6且f(x)在(0,1/3]上是增函数，下面用归纳法证明，
当n=1时，0<a1<1/2，显然成立，当n=2时，0<a2<=1/6<1/(2+1)也成立；
假设当n=k时，0<ak<1/(k+1)也成立，下面证明0<a(k+1)<1/(k+2)即可。
a(k+1)=f(ak)=ak-3/2*ak^2=-3/2(ak^2-2/3ak)=-3/2(ak-1/3)^2+1/6<=1/6，
由于f(x)在(0,1/3]上是增函数，则当0<ak<1/(k+1)时，a(k+1)=f(ak)也是增函数，则有
a(k+1)>f(0)=0,ak=0时，
a(k+1)<f(1/(k+1)),ak=1/(k+1)时,
f(1/(k+1))=1/(k+1)-3/2*1/(k+1)^2=1/(k+1)[1-3/2*1/(k+1)]=1/(k+1)*(2k-1)/[2(k+1)]=(2k-1)/[2(k+1)^2]
(2k-1)/[2(k+1)^2]-1/(k+2)=[(2k-1)(k+2)-2(k+1)^2]/[2(k+1)^2(k+2)]=(-k-4)/[2(k+1)^2(k+2)]<0,
故a(k+1)<f(1/(k+1))<1/(k+2),因此0<a(k+1)<1/(k+2)，即该命题成立。