Answer to Question #260218 in Real Analysis for abc

Question

Check the convergence of the sequence defined by 𝑢𝑛+1 = 1 2 (𝑢𝑛 + 𝑎 𝑢𝑛 ) , 𝑎 > 0. Note that this is the sequence associated with finding the square root of a number 𝑎 > 0 by the Newton’s method.

Answer 1

let the sequence $u_n$ converges to $l$ .

Then,

$u_{n+1}=\frac{1}{2}(u_n+\frac{a}{u_n})\\ l=\frac{1}{2}(l+\frac{a}{l})\\ \text{Solving this equation,we get}\\ l=\frac{1}{2}(\frac{a+l^2}{l})\\ 2l^2=a+l^2\\ l^2=a\\ l=\pm\sqrt{a}\\ \text{Since, it is given a>0. Therefore, }\\l=\sqrt{a}$

Checking hypothesis that the sequence converges, we get the value of sequence limit $\sqrt{a}$

So, we can conclude that this sequence converges.