let $c>0$ and $a_{1} = (c/2)$, $a_{n+1} = \frac{c + a_{n}^2}{2}$. Let $c&gt;0$ 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?
First, $a_{2} = \frac{c + a_{1}^2}{2}&gt;a_1=\frac{c}{2}$, and for $n\geq 2$
$$a_{n+1}-a_n=\frac{c + a_{n}^2}{2}-\frac{c + a_{n-1}^2}{2}=\frac{(a_n-a_{n-1})(a_n+a_{n-1})}{2}&gt;0.$$
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 $$0&lt;c\leq 1.$$
When $c\in(0,1]$, we prove the sequence is bounded. By induction if $0&lt;a_n&lt;1-\sqrt{1-c}$ (you can check that $a_1&lt;1-\sqrt{1-c}$), then 
$$a_{n+1} = \frac{c + a_{n}^2}{2}&lt;\frac{c+(1-\sqrt{1-c})^2}{2}=1-\sqrt{1-c}.$$
So when $c\in(0,1]$, the limit is $1-\sqrt{1-c}$.
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