求各位大大,一道关于数列的题.

an=t^2+(1-t)Sn;a(n-1)=t^2+(1-t)S(n-1);
两式相减an-a(n-1)=an-an*t;推出an/a(n-1)=1/t;为等比数列
又a1=t^2+(1-t)a1;推出a1=t;所以an=t^(2-n);
bn=2/(3-logtan)=2/(3-(2-n))=2/(1+n);
bn*b(n+1)=4/(1+n)*(2+n)=4(1/(1+n)-1/(2+n));
Pn=4(1/2-1/3+1/3-1/4.-1/(2+n))=4(1/2-1/(2+n))<2;