级数∑∞n=11/[(2n-1)(2n+1)]的和S=____.
1/2 分析：由于1/(2n-1)(2n+1)=1/2[1/(2n-1)-1/(2n+1)]1/(1•5)+1/(3•5)+1/(5•7)+…1/(2n-1)(2n+1) =1/2[1-1/3+1/3-1/5+1/5-1/7+…+1/(2n-1)-1/(2n+1)] =1/2[1-1/(2n+1)] 所以limn→∞2/n[1-1/(2n+1))=1/2．
级数∑∞n=11/[(2n-1)(2n+1)]的和S=____.
1/2 分析：由于1/(2n-1)(2n+1)=1/2[1/(2n-1)-1/(2n+1)]1/(1•5)+1/(3•5)+1/(5•7)+…1/(2n-1)(2n+1) =1/2[1-1/3+1/3-1/5+1/5-1/7+…+1/(2n-1)-1/(2n+1)] =1/2[1-1/(2n+1)] 所以limn→∞2/n[1-1/(2n+1))=1/2．
