(1)lgnlg(n+2)<lg2(n+1);
(2)logn+1n<logn+2(n+1).
(1)lgnlg(n+2)<lg2(n+1);
(2)logn+1n<logn+2(n+1).
证明:(1)∵n>1,∴lgn>0,lg(n+2)>0.
∴lgn·lg(n+2)≤[]2=
lg2(n2+2n)<
lg2(n2+2n+1)=lg2(n+1).
∴lgnlg(n+2)<lg2(n+1).
(2)由(1)知lgnlg(n+2)<lg(n+1)lg(n+1),
∴,即logn+1n<logn+2(n+1).