Does the sequence $x_{n+1} = \frac{1}{2} ( (x_n)^2 + c ) $ diverge? Does this recursive sequence defined by the relation 

$x_{n+1} = \frac{1}{2} ( (x_n)^2 + c ) $

diverge. Given that we know that , $x_0 &gt; 1$ and c lies in the open interval (0,1)

If yes then please how ? If not then also please proof.

My attempt : My initial attempt was to assume that the sequence diverges , given that it is strictly  increasing. However, I realised that I could not actually proof that the series is going towards infinity. ( After one of my friends said that just saying that the series is monotonically increasing won't be enough ) 

So far my attempt to prove that the sequence goes to infinity has been to find the general term , and then show that as n tends to infinity the function also tends to infinity. However i could not find the general term . 

So I tried another approach too , which was basically trying to find a  sequence $b_n$such that each term of the new sequence is smaller than the corresponding  term of the original sequence  {$a_n$} and then to show that $b_n$ diverges . 

However, I haven't found such a sequence yet !.

Any help on whether I am proceeding in the right direction , and proof of whether the sequence converges or diverges will be appreciated.
If the sequence converges to some limit $L$, then we must have
$$L = \frac{1}{2}(L^2 + c)$$
so
$$L^2 - 2L  = -c$$
Completing the square on the left hand side, this becomes
$$(L - 1)^2 = 1-c$$
and therefore
$$L = 1 \pm\sqrt{1-c} &lt;2$$
where the inequality holds because $0 &lt; c &lt; 1$.



Now consider $x_0 = 3$ and $c=1/2$. If $x_n \geq 3$ then
$$x_{n+1} = \frac{1}{2}\left(x_n^2 + \frac{1}{2}\right) \geq \frac{1}{2}\left(3^2 + \frac{1}{2}\right)= \frac{19}{4} \geq 3$$
so by induction we see that $x_n \geq 3$ for all $n$. Thus the sequence cannot converge to a limit smaller than $3$. But as shown above, if it does converge to a limit $L$, then $L &lt; 2$. This means that the sequence diverges.

On the other hand, as pointed out by Logan Maingi in the comments, if $x_0 = 3/2$ and $c = 1/4$, then $x_n$ is strictly decreasing and bounded below by $0$, so it converges.

So, if all we know is that $x_0 &gt; 1$ and $c \in (0,1)$, that is insufficient information to decide whether $x_n$ converges or diverges. It depends on the specific values of $x_0$ and $c$, and this dependence is shown precisely in the answer by mathguy.
Let $L = \lim_{n\to\infty}x_n$. Then $\lim_{n\to\infty}x_{n+1} = L$ as well. Therefore we have
    $$1 =\lim_{n\to\infty}\frac{x_n}{x_{n+1}} = \lim_{n\to\infty}\frac{x_n}{\frac{1}{2}((x_n)^2 +c)} = \frac{L}{\frac{1}{2}(L^2 + c)}$$

You should be able to solve for $L$, the solution will depend on $c$.