Let $0 < a \leq 1$ and $s_1 = a/2$ , $2s_{n+1} = s_n^2 + a$ , Then how to show that the sequence is convergent.

Question

Let $0 < a \leq 1$ and $s_1 = a/2$ , $2s_{n+1} = s_n^2 + a$ , Then how to show that the sequence is convergent.

My Try : I have tried to find out $s_{n+1} - s_n$ and try to understand the nature of the sequence. I got that if $a >1$ then the sequence would be monotonically increasing. But we can not say anything when $a<1$. Can anyone please help me out?

Answer 1

In the case $0<a<1$ you can proceed in two steps:

1) Prove by induction that $s_n < 1- \sqrt{1-a}$.

Namely, this is easily proved for $n = 1$; if we assume that the inequality holds for some $n$, then $$ s_{n+1} = \frac{s_n^2 + a}{2} < \frac{(1-\sqrt{1-a})^2+a}{2} = 1 - \sqrt{1-a}. $$

2) Show that the sequence $(s_n)$ is monotone increasing.

Indeed $$ s_{n+1} - s_n = \frac{s_n^2 -2s_n + a}{2} > 0, $$ since $x^2 - 2x + a > 0$ for $x < 1-\sqrt{1-a}$.

Answer 2

Bounded:

Let's prove by induction that \begin{equation} 0 < s_{n+1} < 1 \end{equation} The base case is obvious. Now assume $0 < s_n < 1$, hence \begin{equation} 0 < s_n^2 < 1 \end{equation} but \begin{equation} 0 < a < 1 \end{equation} So \begin{equation} 0 < s_n^2 + a < 2 \end{equation} Therefore \begin{equation} 0 < s_{n+1} = 0.5(s_n^2 + a) < 1 \end{equation}

Monotone:

Now let's prove by induction that $s_{n}$ is increasing. $s_1 < s_2$ is obvious. Then, assuming $s_{n-1} < s_n$, we have \begin{equation} s_{n-1}^2 + a < s_n^2 + a \end{equation} We get \begin{equation} s_{n} < s_{n+1} \end{equation} So, the recursion must converge.

Answer 3

First, $s_{2} = \frac{a + s_{1}^2}{2}>s_1=\frac{a}{2}$, and for $n\geq 2$ $$s_{n+1}-s_n=\frac{a + s_{n}^2}{2}-\frac{a + s_{n-1}^2}{2}=\frac{(s_n-s_{n-1})(s_n+s_{n-1})}{2}>0.$$ So the sequence$\{s_n\}$ is strictly increasing. If the sequence is bounded, then the limit exists, suppose the limit exists and is $x$, then it must satisfy $$x=\frac{a+x^2}{2}.$$ If this equation has solution($x=1\pm\sqrt{1-a}$), it must be $$0<a\leq 1.$$ When $a\in(0,1]$, we prove the sequence is bounded. By induction if $0<s_n<1-\sqrt{1-c}$ (you can check that $s_1<1-\sqrt{1-a}$), then $$s_{n+1} = \frac{a + s_{n}^2}{2}<\frac{a+(1-\sqrt{1-a})^2}{2}=1-\sqrt{1-a}.$$ So when $a\in(0,1]$, the limit is $1-\sqrt{1-a}$.