假设收敛,可以设a=limsinn,则limsin(n+2)=a.而sin(n+2)-sinn=2cos(n+1)sin1,得lim2cos(n+1)sin1=a-a=0,则limcos(n+1)=0,limcosn=0.则a=limsinn=lim√(1-cos^2 n)=1.又 sin2n=2sinncosn,两边取极限,得a=2a×0,矛盾.所...