TITLE: If $(u_n)$ is bounded and $\lim u_n^2+u_n-u_{n+1}=0$ then $u_n \to 0$
QUESTION [6 upvotes]: Let $(u_n)$ be a real bounded sequence. Suppose that $\lim\limits_{n \to \infty} (u_n^2+u_n-u_{n+1})=0$. Prove that $u_n \to 0$.
I was able to develop a prove, looking at the map $x \mapsto x^2+x$ and proving that $0$ is the only possible limit point of $(u_n)$. But the proof has several cases... Not something very straightforward.
Do your have any idea of something simple?

REPLY [2 votes]: Suppose $u_n$ did not go to zero. Then since the sequence is bounded, $L = \limsup_{n \rightarrow \infty} |u_n| > 0$. Let $\{n_k\}_{k = 1}^{\infty}$ be such that $\lim_{k \rightarrow \infty} u_{n_k} = L$ or $-L$. First we consider the $L$ case. Then one has
$$\lim_{k \rightarrow \infty} u_{n_k + 1}= \lim_{k \rightarrow \infty} (u_{n_k} +  u_{n_k}^2) + \lim_{k \rightarrow \infty}
(u_{n_k +1} - (u_{n_k} +  u_{n_k}^2))$$
$$= L + L^2 - 0$$
$$= L + L^2$$
Since $L + L^2 > L$, this contradicts that $L$ is the limsup of the absolute values.
Now suppose the subsequence has limit $-L$. This time we observe that
 $$-L = \lim_{k \rightarrow \infty} u_{n_k} = \lim_{k \rightarrow \infty} (u_{n_k - 1} +  u_{n_{k}-1}^2) + \lim_{k \rightarrow \infty}
(u_{n_{k}} - (u_{n_{k}-1} +  u_{n_{k}-1}^2))$$
$$= \lim_{k \rightarrow \infty} (u_{n_k-1} +  u_{n_k-1}^2)$$
Passing to a convergent subsequence $u_{n_k-1}$, if the limit of the subsequence is denoted by $M$ then by the above we have
$$M + M^2 = -L$$
Thus $M < -L$, contradicting that the negative number $L$ is the limsup of the absolute values of the sequence. 
Thus regardless of which case we are in, we get a contradiction, so the limit must be zero.