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

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

• 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

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 $$s_{n+1}-s_n=\frac{3-s_n^2}{2s_n}<0.$$