(1)若a=0时,a1=2,an+1=, ∴an+12=an且an>0. 两边取对数,得2lgan+1=lgan, ∵lga1=lg2, ∴数列{lgan}是以lg2为首项,为公比的等比数列, ∴lgan=()n-1lg2,即an=221-n; (2)由an+1=,得an+12=an+a,① 当n≥2时,=an-1+a,② ①-②,得(an+1+an)(an+1-an)=an-an-1, 由已知可得an>0,∴an+1-an与an-an-1同号, ∵a2=,且a>0,∴-=(a+2)2-(2a+2)=a2+2a+2>0恒成立, ∴a2-a1<0,则an+1-an<0. ∵bn=|an+1-an|,∴bn=-(an+1-an), ∴Sn=-[(a2-a1)+(a3-a2)+…+(an+1-an)]=-(an+1-a1)=a1-an+1<a1. |