正数数列n中,Sn为其前n项和，且Sn=1/2(An+1/An),猜想An

A1 = S1 = (1/2)(A1 + 1/A1)
A1 = 1

A1 + A2 = 1 + A2 = S2 = (1/2)(A2 + 1/A2)
A2 = √2 -1

A1 + A2 + A3 = √2 + A3 = (1/2)(A3 + 1/A3)
A3 = √3 - √2

猜想
An = √n - √(n-1)

如成立, 则
1/An = √n + √(n-1)

An + 1/An = 2√n

Sn = √n

A1, A2 A3 时已经成立

假设 n = k 时 以上各式成立

当 n = k+1 时
S(k+1) = (1/2) (A<k+1> + 1/A<k+1>
Sk + A<k+1> = (1/2)(A<k+1> + 1/A<k+1>)
√k + A<k+1> = (1/2)(A<k+1> + 1/A<k+1>)
2√k + 2A<k+1> = A<k+1> + 1/A<k+1>
2√k + A<k+1> = 1/A<k+1>
(A<k+1> )^2 + 2√k * A<k+1> = 1
(A<k+1> )^2 + 2√k * A<k+1> + (√k)^2 = k + 1
[A<k+1> - √k]^2 = √(k+1)
数列为正数列, 所以
A<k+1> = √(k+1) - √k

因此 , 当 n =k 成立时候, n = k+1 也成立.
所以对于一切 n 成立

An = √n - √(n-1)

计算出a1=1,a2=√2-1,a3=√3-√2
猜想an=√n-√(n-1)
用