Whether the sequence following is convergent? $c<-3$ is a real number, $\{x_n\}$ is a sequence of real number, $\displaystyle x_1=\frac{c}{2}$, and $\displaystyle x_{n+1}=\frac{c}{2}+\frac{x_n^2}{2}$. Whether $\{x_n\}$ is a convergent sequence?
 A: Any limit of an iterative sequence like this must be a fixed point of the iteration function. So if there is a limit for your sequence, it must satisfy $x=\frac{c+x^2}{2}$. These fixed points are stable if $|f'(x)|<1$, unstable if $|f'(x)|>1$. In the first case, the iteration will converge for initial conditions close enough to the fixed point. In the second case, the iteration will never converge to that fixed point, unless the initial condition is exactly at the fixed point, of course. 
So for your problem, solve $x=\frac{c+x^2}{2}$; substitute the solutions into $f'$; check the condition I wrote. If you get $|f'(x)|>1$ then you are done, you will not have convergence. (Note that the initial condition in your case is never a fixed point, as is easy to check.) If you get $|f'(x)|<1$ then you will have convergence for some initial conditions, but possibly not for the specific initial condition that you have. In this case there is more work to be done. If you get $|f'(x)|=1$ then you have made no progress at all.
A: So, following up on what was previously posted, it appears that the answer is no, if you insist on $c < 3$, because if you solve for the fixed point, you get $x = 1 \pm \sqrt{1-c}$. The only place you get a single number for $x$ is when $c$ is $+1$. Starting with that value of $c$ the sequence very slowly (appears to) converge to $1$ (as you would expect since you are taking $c/2$ and adding half of a number near to $1$ to it). If you let $c = -4$, the sequence alternates between $-2$ and $0$ (forever). Interestingly, if you start with $c = -8$, the first term is $-4$ and every term thereafter is $4$. But, also interestingly, if you start with $c = 1.001$ or $-8.001$, it blows up very quickly.
