(1) Multiply the two formulas according to the hypothesis to get a (n+1) b (n+1) = AnBN, A 1b 1 = 4, AnBN = 4, and use a (n+60). =2, if n & gt=2; And A 1=4, then Ann's proposition is proved; AnBn=4, add a & gt2st.0 <; Bn<2, the first question is over.

(2) According to the title, A(n+ 1)=(An square +4)/2An, and the fixed point is plus or minus 2, so (An+2/An-2) constitutes a square recursive sequence, which shows that (an+2/an-2) = 3 ∧ (2 ?).

(3) let Kn=3∧(2∧(n- 1)), then we can know that An=2(Kn+ 1)/(Kn- 1), bn = 2 (kn-/kloc-0).

The formula to be proved is ∑ (kn+1)/(kn-1)/(kn+1) < 2n+4/3, that is ∑1/kn-/.

So ∑ 1/KN- 1

Complete the certificate.