let $c>0$ and $a_{1} = (c/2)$, $a_{n+1} = \frac{c + a_{n}^2}{2}$.

Question

Let $c>0$ and $a_{1} = \frac{c}2$, $a_{n+1} = \frac{c + a_{n}^2}{2}$. Determine all $c$ for which the sequence converges . For such $c$ find $\lim_{n\to\infty} a_{n}$.

Could anyone help me please in solving this?

Answer 1

First, $a_{2} = \frac{c + a_{1}^2}{2}>a_1=\frac{c}{2}$, and for $n\geq 2$ $$a_{n+1}-a_n=\frac{c + a_{n}^2}{2}-\frac{c + a_{n-1}^2}{2}=\frac{(a_n-a_{n-1})(a_n+a_{n-1})}{2}>0.$$ So the sequence$\{a_n\}$ is strictly increasing. If the sequence is bounded, then the limit exists, suppose its limit is $a$, then it must satisfy $$a=\frac{c+a^2}{2}.$$ If this equation has solution($a=1\pm\sqrt{1-c}$), it must be $$0<c\leq 1.$$ When $c\in(0,1]$, we prove the sequence is bounded. By induction if $0<a_n<1-\sqrt{1-c}$ (you can check that $a_1<1-\sqrt{1-c}$), then $$a_{n+1} = \frac{c + a_{n}^2}{2}<\frac{c+(1-\sqrt{1-c})^2}{2}=1-\sqrt{1-c}.$$ So when $c\in(0,1]$, the limit is $1-\sqrt{1-c}$.

Answer 2

Hint: Solve $$l=\frac{c+l^2}{2}$$

to get a list of possible values of the limit (I'll leave it to you to see why this is true)

Edit: Look at the comments for mroe hints

Answer 3

Using the AM-GM inequality we get $$a_{n+1} = \frac{c+a_n^2}2 \ge \sqrt{ca_n^2} = \sqrt{c}a_n$$

so if $a_n\to L \ge 0$, we get $L \ge \sqrt{c} L$, or $c \le 1$.

Also letting $n\to\infty$ in $a_{n+1} = \frac{c+a_n^2}2$ gives $L = \frac{c+L^2}2$ so $L = 1 \pm \sqrt{1-c}$. Since $c \le 1$, inductively we can prove that $a_n \le 1, \forall n \in \mathbb{N}$ so $L \le 1$. Therefore $L = 1 -\sqrt{1-c}$ is the only candidate for the limit.

For any $c \in \langle 0,1]$ then inductively show that the sequence $(a_n)_n$ increases so it indeed converges to $1 -\sqrt{1-c}$.