(1)an+1-an=- =(n≥2), 上式表明an+1-an与an-an-1同号, ∴an+1-an,an-an-1,an-1-an-2,…,a2-a1同号, ∵a>2, ∴a2-a-2=(a-2)(a+1)>0, ∴a2>a+2, ∴a2=<a,a2-a1<0. ∴an+1-an<0, 故an+1<an. (2)∵an+1=bn+1+ = =, bn+12+=bn+, bn+14-(bn+)bn+1 2+1=0, 注意到bn>1, f(x)=x+(x>0),f′(x)=1->0, ∴f(x)在x>1时为增函数,而f(bn+12)=f(bn), ∴bn+12=bn, ∴2lgbn+1=lgbn, ∴=, ∴数列{lgbn}是等比数列, 当a1=b1+=,b1=,lgb1=lg, lgbn=()n-1•lg=()n•lg2, ∴bn=2()n, an=bn+=2()n+2-()n. (3)∵当n≥2时,an-2=-2=, 上式表明:an-2与an-1-2同号,对一切n≥2成立, ∴an-2,an-1-2,…,a2-2,a1-2同号, 而a1-2>0, ∴an-2>0,an-1-2>0, ∵n≥2时,an-2=<=, ∴<, ∴•…• =<()n-1, ∴0<an-2<(a1-2)•()n-1, 当a1=2011,n=12时, a12-2=(2011-2)×()12-1=<=<, ∴a12<2+, ∵an>an+1, ∴当n≥12时,2<an<2+恒成立. |