Find the limit if it exists of $S_{n+1} = \frac{1}{2}(S_n +\frac{A}{S_n})$ [duplicate]

Suppose that $S_0$ and A are positive numbers, let $$S_{n+1} = \frac{1}{2}\left(S_n +\frac{A}{S_n}\right)$$ with $n \geq 0 $.

(a)Show that $S_{n+1} \geq \sqrt{A} $ if $n \geq 0$

(b)Show that $S_{n+1} \leq S_n $ , if $n \geq 1$

(c) Show that $s= \lim\limits_{n \rightarrow \infty} S_n$ exists

(d) find s

(a) Show that $S_{n+1} \geq \sqrt{A} $ if $n \geq 0$

Given $$P_n: S_{n+1} = \frac{1}{2}\left(S_n +\frac{A}{S_n}\right) \geq \sqrt{A}$$

$$P_0: S_{1} = \frac{1}{2}\left(S_0 +\frac{A}{S_0}\right) \geq \sqrt{A} $$

We assume that $P_n$ is true

$$P_{n+1}: S_{n+2}= \frac{1}{2}\left(S_{n+1} +\frac{A}{S_{n+1}}\right)$$ by assumption $$S_{n+2}= \frac{1}{2}\left(S_{n+1}\left(1 +\frac{A}{(S_{n+1})^2}\right)\right) \geq \frac{1}{2}\left(\sqrt{A}\left(1 +\frac{A}{(\sqrt{A})^2}\right)\right)$$ $$ S_{n+2}= \frac{1}{2}\left(S_{n+1} +\frac{A}{S_{n+1}}\right) \geq \sqrt{A} $$

It follows that $S_{n+1} \geq \sqrt{A} $

(b) Show that $S_{n+1} \leq S_n $ , if $n \geq 1$

$$S_{n+1} \leq S_n$$ $$\frac{1}{2}\left(S_n +\frac{A}{S_n}\right) \leq S_n $$ Dividing by $S_n$ $$\frac{1}{2}\left(1 +\frac{A}{S_n^2}\right) \leq 1 $$ $$\frac{A}{2S_n^2} \leq \frac{1}{2}$$ $$A \leq S_n^2$$ $$S_n \geq \sqrt{A}$$ As $S_{n+1} \leq S_n$ yields a true statement, it follows $S_{n+1} \leq S_n$ is true.

(c) Show that $s= \lim\limits_{n \rightarrow \infty} S_n$ exists

Since $S_{n+1} \leq S_n$, the sequence is non-increasing,

using the non-increasing theorem stating that

if $\{S_n\}$ is non-increasing then $$\lim\limits_{n \rightarrow > \infty} S_n = \inf\{S_n\} $$

(d) find s

Is the argumentation in (a) and (b) appropriate? Also, I have to admit I m getting less confident in my argumentation (c) and (d). How to proceed in (c) and (d)? Much appreciated for your input or help.

Solution 1:

Hint You can simplify a) by applying AM-GM, obviously showing that each $S_n>0$ which is easy to show $$\frac{1}{2}\left(S_n+\frac{A}{S_n}\right)\geq \sqrt{S_n \cdot \frac{A}{S_n}}=\sqrt{A}$$ For c) more details here and you already proved the sequence is decreasing and bounded below in b) and a).

For d) as Eugen suggested, with a), b) and c) you are confident that the sequence is converging so: $$\lim_{n \rightarrow \infty} S_{n+1}= \lim_{n \rightarrow \infty} \frac{1}{2}\left(S_n+\frac{A}{S_n}\right) \Rightarrow s=\frac{1}{2}\left(\lim_{n \rightarrow \infty}S_n+\frac{A}{\lim\limits_{n \rightarrow \infty} S_n}\right)\Rightarrow \\ s=\frac{1}{2}\left(s+\frac{A}{s}\right)$$ obviously $s \geq \sqrt{A}>0$, thus $$2s^2=s^2+A \Rightarrow s=\sqrt{A}$$

Solution 2:

a) $\frac{1}{2}(S_{n+1}(1 +\frac{A}{(S_{n+1})^2}) \geq \frac{1}{2}(\sqrt{A}(1 +\frac{A}{(\sqrt{A})^2})$ is not necessarily true for $S_{n+1} \ge \sqrt{A}$ because $\frac{A}{(S_{n+1})^2} \le \frac{A}{(\sqrt{A})^2}$. You can prove $\frac{1}{2}(S_{n+1} +\frac{A}{S_{n+1}}) \ge \sqrt{A}$ simply transforming it into $(S_{n+1} - \sqrt{A})^2 \ge 0$

d) To find $s$ just replace $S_{n+1}$ and $S_{n}$ in the recurrence formula with $s$ and solve the resulting equation