By assuming a1= s1>; 1, so a 1=2,

And an+1= sn+1-sn =-(an+1+1)-(an+1)-(an+2),

get(an+ 1+an)(an+ 1-an-3)= 0，

That is, an+ 1-an-3=0 or an+ 1=-an because an >；; 0, so an+ 1=-an is invalid, so it is discarded.

So an+ 1-an=3, so {an} is a arithmetic progression with a tolerance of 3 and the first term of 2.

So the general term of {an} is an=3n- 1.

(2) Proof: Through comparison. It can be solved by an (- 1) = 1.

bn = log2( 1+-)= log2-；

Therefore TN = b1+B2+...+BN = log2 (-■ ...-).

Therefore 3tn+1-log2 (an+3) = log2 (-■…-) 3 ■.

Let f (n) = (-■…-) 3 ■,

Then -=-(-) 3 =-.

Because (3n+3) 3-(3n+5) (3n+2) 2 = 9n+7 >; 0, so f(n+ 1) > female (noun).