(1)∵即,∴公差d=-1. 且首项为,故是等差数列. (2)∵,∴. 设f(x)=x-ln(x+1),(x>0),则,f(x)在(0,+∞)↑,且f(x)在[0,+∞)上连续,∴f(x)>f(0)=0,∴x>0时x>ln(x+1), ∴,即. ∴an<1-ln(n+1)+lnn,∴Sn<(1-ln2+ln1)+(1-ln3+ln2)+…+[1-ln(n+1)+lnn]=n-ln(n+1)故Sn<n-ln(n+1). (3)∵,∴,当时,则,∴, 即n≥4;又当时,则,即n≤3,因此得b1<b2<b3<b4>b5>b6>…,又∵b1=0,n≥2时,bn>0,∴0≤bn≤b4.∴对任意正整数n、m,都有 |