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. 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?
 A: 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 $x < 1-\sqrt{1-a}$.
A: 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.
A: 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 $1-\sqrt{1-a}$.
