Solution:

(2) By 4Sn=an？ +2an

Get: 4S(n- 1)=a(n- 1)? +2a(n- 1)

Where n≥2 and n is an integer.

Subtract the two expressions to get:

4an=an？ -a(n- 1)？ +2an-2a(n- 1)

that is

Ann? -a(n- 1)？ -2an-2a(n- 1)=0

The left factor of the formula is decomposed into:

[an+a(n- 1)][an-a(n- 1)-2]= 0

Since {an} are all positive numbers, an+a(n- 1)>0.

So there must be:

an-a(n- 1)-2=0

that is

an=a(n- 1)+2

Therefore, {an} is a arithmetic progression, the first term is 2, and the tolerance is 2, then

an=2+2(n- 1)=2n

(3)

1/a 1？ + 1/a2？ +...+ 1/an？

= 1/2? + 1/4? +...+ 1/(2n)？

& lt 1/2? + 1/4 2+ 1/6 4+...+ 1/2n(2n-2)

= 1/4+[ 1/2- 1/4+ 1/4- 1/6+...+ 1/(2n-4)- 1/(2n-2)+ 1/(2n-2)- 1/2n)/2

= 1/4+ 1/4- 1/4n = 1/2- 1/4n

∫n≥ 1

∴ 1/a 1？ + 1/a2？ +...+ 1/an？ & lt 1/4+ 1/4- 1/4n & lt； 1/2