（1）∵(1-1/2'2)(1-1/3'2)...(1-1/n'2)
=[(1+1/2)(1-1/2)][(1+1/3)(1-1/3)].[(1+1/n)(1-1/n)]
=[(1+1/2)(1+1/3).(1+1/n)][(1-1/2)(1-1/3).(1-1/n)]
=[(3/2)(4/3).((n+1)/n)][(1/2)(2/3).((n-1)/n)]
=[(n+1)/2](1/n)
=(1+1/n)/2
∴原式=lim(n->∞)[(1+1/n)/2]
=(1+0)/2
=1/2
（2）∵设Sn=1/2+3/2^2+5/2^3+...+(2n-1)/2^n.(1)
1/2Sn=1/2^2+3/2^3+5/2^4+...+(2n-1)/2^(n+1).(2)
(1)-(2)得1/2Sn=1/2+2/2^2+2/2^3+...+2/2^n-(2n-1)/2^(n+1)
=1/2-(2n-1)/2^(n+1)+[1/2+2/2^2+...+2/2^(n-1)]
=1/2-(2n-1)/2^(n+1)+1-1/2^(n-1)
=3/2-(2n+3)/2^(n+1)
∴Sn=3-(2n+3)/2^n
故原式=lim(n->∞)[3-(2n+3)/2^n]
=3-lim(n->∞)[(2n+3)/2^n]
=3-0 (∞/∞型,应用罗比达法则)
=3