(1)依题意：2lga2=lga1+1+lga4,即a2^2=10a1*a4
(a1*q)^2=10a1*a1*q^3,得q=1/10
a1a2a3=1,即a1*a1q*a1q^2=1,an>0,可求得a1=10
所以an=10*(1/10)^(n-1)=10^(2-n)
(2)bn=1/n（3-lgan）,=1/n(3-2+n)=1/n-1/(n+1)
b1=1/1-1/2;
b2=1/2-1/3
b3=1/3-1/4
…………
bn=1/n - 1/(n+1)
Tn=1/1-1/2+1/2-1/3+1/3-1/4+…+1/n - 1/(n+1)
=1-1/(n+1)
1/(n+1)>0
所以Tn