Firstly, it is proved that there is an upper bound, that is, for any n, xn is less than or equal to a constant c.

We prove that xn

1 . x 1 =√2 & lt； 2;

2. let xk < =2, x (k+1) = √ (2+x (k)) <; =√(2+2)=2;

It can be seen that xn < 2；;

It is proved again that xn is monotonically increasing;

I already know xn

√x(n- 1)* x(n- 1)= x(n- 1)； The above derivation is based on x (n- 1).

So xn & gt=x(n- 1), so xn is a monotonically increasing sequence.

It is proved that the xn sequence is monotonically increasing and has an upper bound, so the limit exists.

In fact, the limit of this series is 2, and the calculation limit can be calculated like this.

Let x be the limit of xn, and the limits on both sides of the formula xn=√(2+x(n- 1)) are as follows.

X=√(2+x), the solution is x=2, which means x=2.