设数列{bn}的公差为d,由题意得,∴bn=3n-2 (2)证明:由bn=3n-2知 Sn=loga(1+1)+loga(1+)+…+loga(1+) =loga[(1+1)(1+)…(1+)] 而logabn+1=loga,于是,比较Sn与logabn+1的大小比较(1+1)(1+)… (1+)与的大小. 取n=1,有(1+1)= 取n=2,有(1+1)(1+ 推测:(1+1)(1+)…(1+)> (*) ①当n=1时,已验证(*)式成立. ②假设n=k(k≥1)时(*)式成立,即(1+1)(1+)…(1+)> 则当n=k+1时,
,即当n=k+1时,(*)式成立 由①②知,(*)式对任意正整数n都成立. 于是,当a>1时,Sn>logabn+1,当 0<a<1时,Sn<logabn+1 |