# Monotonic recursive sequence with recursive term in denominator: $s_{n+1} = \frac{1}{2} \left(s_n + \frac{3}{s_n}\right)$ [duplicate]

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I am trying to show that $s_{n+1} = \frac{1}{2} \left(s_n + \frac{3}{s_n}\right)$ is a monotonic decreasing sequence for $n \ge 2$. Currently, my approach using induction is stuck because of the $\mathbf{s_n}$ term appearing in the denominator.

What I have so far:

Calculating a few terms: $s_1 = 1$, $s_2 = 2$, $s_3 = \frac{7}{4} < s_2$. For this to be monotonic decreasing, we have to show $s_{k+2} < s_{k+1}$.

For the base case, we assume for $k$ we have $s_{k+1} < s_k.$

The inductive step: to show $s_{k+2} < s_{k+1}$, i.e. to show $$\color{grey}{s_{k+2} = }\frac{1}{2} \left(s_{k+1} + \frac{3}{s_{k+1}}\right) < \frac{1}{2} \left(s_{k} + \frac{3}{s_{k}}\right) \color{grey}{= {s_{k+1}}_.}$$

In order to show $\dfrac{1}{2} \left(s_{k+1} + \frac{3}{s_{k+1}}\right) < \dfrac{1}{2} \left(s_{k} + \frac{3}{s_{k}}\right)$, I am stuck:

Given, $s_{k+1} < s_k$, I cannot then say $s_{k+1} + \mathbf{\frac{3}{s_{k+1}}} < s_k + \mathbf{\frac{3}{s_{k}}}$, because the $s_k, s_{k+1}$ terms which appears in the denominator may reverse the inequality. Any pointers on how to proceed further?

Disclaimer: I am revising real analysis on my own from the Kenneth Ross book, this is not strictly homework.

## marked as duplicate by rtybase, Namaste, JonMark Perry, Claude Leibovici, Parcly TaxelFeb 28 '18 at 1:34

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• If $s_1=1$ then $s_2=2$. How's this sequence a monotonically decreasing sequence then? – Math Lover Feb 26 '18 at 20:34
• You are applying the Babylonian method for computing $\sqrt{3}$, which is equivalent to Newton's method applied to $f(x)=x^2-3$. – Jack D'Aurizio Feb 26 '18 at 20:42
• This is Newton-Raphson method to solve the equation $x^2-3=0$. – hamam_Abdallah Feb 26 '18 at 20:44
• @MathLover corrected the question in response to your comment – AruniRC Feb 26 '18 at 20:44

## 3 Answers

Presuming $s_1 > 0$, by AM-GM inequality, $$s_{n} = \frac{s_{n-1}+\frac{3}{s_{n-1}}}{2} \ge \sqrt{3}$$ for $n \ge 2$. Also, $$s_{n+1}-s_{n} = \frac{1}{2}\left(\frac{3}{s_n}-s_n\right)\le 0 \iff s_{n} \ge \sqrt{3}$$ for $n \ge 2$.

Assume $s>0$ and notice that for $s\ne\sqrt3$

$$\frac12\left(s+\frac3s\right)>\sqrt3$$ as can be established by computing the minium.

So for $n>1$, we have

$$s_n>\sqrt3\implies s_n^2>3\implies s_n>\frac3{s_n}\implies s_n>\frac12\left(s_n+\frac3{s_n}\right)=s_{n+1}.$$

At the same time this establishes that the sequence is decreasing and bounded below by $\sqrt3$, thus is convergent.

The hint.

Use for all $n\geq2$ $$s_{n+1}-\sqrt3=\frac{(s_n-\sqrt3)^2}{2s_n}>0$$ and $$s_{n+1}-s_n=\frac{3-s_n^2}{2s_n}<0.$$