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 , $x_0 > 1$ 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.  
 A: 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
$$L = 1 \pm\sqrt{1-c} <2$$
where the inequality holds because $0 < c < 1$.

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 $L < 2$. 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 $x_0 > 1$ 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.
A: Let $a = x_0$; then the answer depends on $a$ AND $c$.
Let $t_1 = 1 - \sqrt{1-c}$ and $t_2 = 1 + \sqrt{1-c}$ be the roots of the quadratic equation $x = \frac 1 2 (x^2 + c)$. 
If $a = t_1$ or $a = t_2$ then $x_1 = a = x_0$ and by induction the sequence is CONSTANT and therefore it converges to $a$ (which, recall, is either $1 - \sqrt{1-c}$ or $1 + \sqrt{1-c}$).
If $a > t_2$ then $x_1 = \frac 1 2 (a^2 + c) > a = x_0$, and by induction the sequence is strictly increasing. It then must have limit $+\infty$; if it was bounded, it would have limit $L$ where $L$ must be either $t_1$ or $t_2$, which is not possible because all the terms are $\ge a > t_2$.
If $0 \le a < t_1$, then on the one hand $x_1 = \frac 1 2 (a^2+c) >a = x_0$, and on the other hand $x_1 < t_1$: $x_1 < t_1 \iff a^2 + c < 2(1-\sqrt{1-c})$ which results directly, by substitution, from $a < t_1 = 1 - \sqrt{1-c}$. So in this case, by induction, we have $x_{n+1} > x_n$ while still $x_{n+1} < t_1$. The sequence is strictly increasing and bounded by $t_1$. Since the limit in this case exists and must equal either $t_1$ or $t_2$, we find that the sequence converges to $t_1$.
This leaves the case $a = x_0$ strictly between $t_1$ and $t_2$. With a similar argument (exercise!), the sequence in this case is strictly DECREASING while all the values are still strictly between $t_1$ and $t_2.$ Conclusion: the sequence is convergent and the limit must be $t_1$.
BONUS:  If $c=1$, this argument shows that $x_n \to 1$ if $0 \le x_0 \le 1$ and diverges to $+\infty$ if $x_0 > 1$. If $c > 1$ then the sequence is strictly increasing to $+\infty$ regardless of $x_0$.
A: (Note: This answer was posted prior to the OP's edit adding the condition $0\lt c\lt1$.)
If $c=0$ and $x_0=2$, the sequence definitely does not diverge!
A: @mathguy @bungo
I post this as an answer but it is only a possible track of answer, of a different type, addressing the problem's intrinsic subtilities.
The structure of the recurrence relationship ($x_{n+1}$ expressed as a quadratic expression of $x_n$ and a parameter $c$) has made me remember to the classical logistic map 
$$x_{n+1}=c x_n(1-x_n)$$
with its associated Feigenbaum diagram (chaotic behavior).
See http://mathworld.wolfram.com/LogisticMap.html
or https://en.wikipedia.org/wiki/Logistic_map
My question is: could this problem be treated as - or even shown more or less equivalent to - the logistic map problem ? 
A: 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 $c$.
