If the complex sequence $u_{n+1}=f(u_n)$ has only one limit point, then it converges I would like to prove that if $f:\mathbb{C}\to\mathbb{C}$ is continuous and if a sequence $u$ defined by :
$\forall n\in\mathbb{N},\,u_{n+1}=f(u_n)$
has only one limit point (not sure of the translation of "valeur d'adhérence" in french), then this sequence does converge.
I guess that it's wise to look after a proof that $u$ is a bounded sequence (it is well known that any bounded complex sequence possessing only one limit point is indeed convergent).
EDIT : we suppose, that for some $u_0\in\mathbb{C}$ - (and not for all of them) - the sequence $(u_n)$ has exactly one limit point.
 A: Suppose $u_{\varphi(n)}$ converges to $\ell\in\Bbb{C}$ for some extraction $\varphi$ (i.e. $\varphi:\Bbb{N}\to\Bbb{N}$ increasing). Since $f$ is continuous at $\ell$, $u_{\varphi(n)+1}=f(u_{\varphi(n)})$ converges to $f(\ell)$. Since $\ell$ is the only limit point of $u$, this implies $f(\ell)=\ell$, i.e. $\ell$ is a fixed point of $f$.
Suppose $u_n$ doesn't converge to $\ell$, i.e. $\exists\epsilon>0,\forall N\in\Bbb{N},\exists n\geq N, |u_n-\ell|> \epsilon$. Since $f$ is continuous at $\ell$ there exists $0<\delta\leq\epsilon$ with
$$f\big(D(\ell,\delta)\big)\subset \overline D(\ell,\epsilon)$$
We can define an extraction $\sigma:\Bbb{N}\to\Bbb{N}$ such that
$$\forall n\in\Bbb{N},\quad
\left\{
\begin{array}{lcc}
|u_{\sigma(2n)}-\ell|&<&\delta\\
|u_{\sigma(2n+1)}-\ell|&>&\epsilon
\end{array}
\right.$$
Let us then set
$$\forall n\in\Bbb{N},\quad
\tau(n)=\max\big\{k\in[\![\,\sigma(2n),\sigma(2n+1)\,]\!]\text{ s.t. }|u_k-\ell|<\delta\big\}
$$
Then by construction $\tau$ is an extraction and
$$\forall n\in\Bbb{N},\quad\delta\leq|u_{\tau(n)+1}-\ell|\leq\epsilon.$$ By compactness of the annulus $A_{\delta,\epsilon}=\{z\in\Bbb{C}\mid \delta\leq|z-\ell|\leq\epsilon\}$, the sequence $u_{\tau(n)+1}$ has a limit point, which contradicts the hypothesis. Thus $(u_n)_n$ converges.
