If $(u_n)$ is bounded and $\lim u_n^2+u_n-u_{n+1}=0$ then $u_n \to 0$ 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?
Suppose $u_n$ did not go to zero. Then since the sequence is bounded, $L = \limsup_{n \rightarrow \infty} |u_n| &gt; 0$. 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 $L + L^2 &gt; L$, 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 $M &lt; -L$, 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.
suppose that $u_{n}$ doesn't converge to zero. So there is a positive number $r$ for which we have $\lvert u_{n} \rvert$ > r, for each n. Define $f(n)= u_{n}^2 + u_{n}- u_{n+1}$ and consider $f(n)+f(n+1)+...f(n+k)= u_{n}^2+u_{n+1}^2+...+u_{n+k}^2+u_{n}-u_{n+k+1}$. Since $u_{n}$ is bounded, there is a positive number M so that we have $\lvert u_{n} \rvert$ &lt; M. Now we can see:
$\lvert f(n)+f(n+1)+...f(n+k) \rvert$ > $\lvert u_{n}^2+u_{n+1}^2+...+u_{n+k}^2 \rvert$ - $\lvert u_{n} \rvert$ - $\lvert u_{n+k+1} \rvert$ > $kr^2-2M$. Choose $k$ big enough so that $kr^2-2M=A&gt;0$. On the other hand
lim $\lvert f(n)+f(n+1)+...f(n+k) \rvert=0$ according to the assumption when n goes to infinite which contradicts to $\lvert f(n)+f(n+1)+...f(n+k) \rvert$ >$A$ >$0$.
( I showed both converging and converging to zero at the same time )