When n =1:2s1= a1+1.

a 1= 1/a 1

a 1？ = 1, then A 1 = 1.

∫ series {an} is a positive number series.

∴a 1= 1

When n≥2: Sn-S(n- 1)=an.

Then 2sn = sn-s (n-1)+1[sn-s (n-1)]

Sn+S(n- 1)= 1/[Sn-S(n- 1)]

Multiply both sides by Sn-S(n- 1):

Sn？ - S(n- 1)？ = 1

∴ series {Sn? } is the arithmetic progression with the first term of 1 and the tolerance of 1.

Then Sn? = 1+(n- 1)？ 1=n

∴Sn=√n

∴an=sn-s(n- 1)=√n-√n- 1，(n≥2)

Then when n= 1, a1= √1-√1=1.

∴an=√n - √n- 1，(n∈N+)