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I am trying to prove the limit of the following sequence is $\sqrt{x}$. I'm stuck on the hint. I don't how they got to the hint.

Problem
Suppose that $x>0$. Define a sequence $S_n$ by $s_1 = k$ and $s_{n+1} = \frac{s_n^2 + x}{2 s_n}$ for $n \in \mathbb{N}$.

Prove that for any $k>0$, $\lim_{n\to\infty} s_n = \sqrt{x}$.

The hint: Show that $(s_{n+1})^2 - x = \frac{(s_n^2 - x)^2} {4s_n^2} \ge 0$, so that $s_n \ge \sqrt{x}$ for $n \ge 2$.

What do I need to do to get to hint? It seems they squared both sides? But, there's an extra $-x$ on the left, and the $+x$ became $-x$ on the right. I am very confused.

user13985
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  • I don't understand the answer in the "similar question". I think the answer below is straight forward to understand. – user13985 Jul 15 '20 at 02:22

1 Answers1

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Start with $s_{n+1} = \frac{s_n^2+x}{2s_n}$ and square both sides: $$ s_{n+1}^2 = \frac{(s_n^4 +2s_n^2 x+x^2)}{4s_n^2} $$ $$ s_{n+1}^2 -x= \frac{(s_n^4 +2s_n^2 x+x^2)}{4s_n^2}-\frac{4s_n^2 x}{4s_n^2} $$ $$ s_{n+1}^2 -x= \frac{(s_n^4 -2s_n^2 x+x^2)}{4s_n^2}=\frac{(s_n^2-x)^2}{4s_n^2} $$Then for $n\geq 1$, $s_{n+1}^2-x\geq 0$ (since the RHS is thing$^2$). This means the sequence $s_n$ is bounded below for $n\geq 2$ by $\sqrt{x}$. If you can then show it is decreasing, you can assert the limit exists, i.e. $\lim_{n\to\infty} s_n= S$, and then use the recurrence to solve it (you should get $S=\sqrt{x}$).

Integrand
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  • Why do we start with square both sides? – user13985 Jul 14 '20 at 16:04
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    Bad answer: because it works. Better answer: the intuition in this problem involves $\sqrt{x}$, so squaring both sides is a plausible way to introduce $x$. Remember our goal is to show $s_n$ is bounded below and decreasing. – Integrand Jul 14 '20 at 16:06
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    For boundedness, I've been using induction by coming up with a number, and show it works for all n. Could that have worked here? – user13985 Jul 14 '20 at 16:26