1.当k>1时,(k-1)^2>3/4<==>4k<(2k-1)^2<==>1/(2k-1)<1/(2√k).2.当k>1时,bk=1/√(ak*a(k+1))=1/√((2k-1)*(2k+1))<<1/(2k-1)<1/(2√k)<<1/(√k+√(k-1))=√k-√(k-1).3.b1=1/√3Sn=b1+ 展开
1.当k>1时,(k-1)^2>3/4<==>4k<(2k-1)^2<==>1/(2k-1)<1/(2√k).2.当k>1时,bk=1/√(ak*a(k+1))=1/√((2k-1)*(2k+1))<<1/(2k-1)<1/(2√k)<<1/(√k+√(k-1))=√k-√(k-1).3.b1=1/√3Sn=b1+b2+..+b(n)<b1+[√2-√(2-1)]+..[√n-√(n-1)]==1/√3+√n-1<√n 收起
