等比数列{an}为2的（n-1）次方,bn=2（log2（an）+1）（n是正整数）.

因为an=2^(n-1),所以bn=2(log2(an)+1)=2[((n-1)+1]=2n,所以(bn+1)/bn=(2n+1)/2n,又因为(2n+1)^2=4n^2+4n+1>4n^2+4n=4n(n+1),所以(2n+1)^2/4n^2>(n+1)/n,(2n+1)/2n>√(n+1)/√n.所以[(b1+1)/b1]*……[(bn+1)/bn=3/2*5...