a1=2,a(n+1)=1/2(an+2/an),n=1,2,3…… Prove LIM (an) = root 2, n tends to infinity

a1=2,a(n+1)=1/2(an+2/an),n=1,2,3…… Prove LIM (an) = root 2, n tends to infinity

Let Liman = x, then Lima (n + 1) = X
Take Lima (n + 1) = Lim [1 / 2 (an + 2 / an)] on both sides,
X = 1 / 2 (x + 2 / x)
That is, x ^ 2 = 2,
So Liman > 0
So Liman = √ 2

Prove Lim an = a, then LIM (an) ^ 2 = a

Lim an = a, LIM (an) ^ 2 = a ^ 2. This is the algorithm of limit product

If the first n terms of sequence {an} and Sn = 2n + n-1, then its general term formula is

Because Sn = 2n ^ 2 + n-1
So an = Sn - S (n-1)
= 2n^2 + n - 1 - 2(n-1)^2 - (n-1) + 1
= 4n - 1
When n = 1, A1 = S1 = 2 + 1-1 = 2
So the general formula is:
An = 2 (when n = 1)
4N-1 (when n > 1)