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?
 A: 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.
A: Let $L=\limsup u_n$, which exists since $u_n$ is bounded. It suffices to show that $L=0$ since then we can also show that $\liminf u_n=0$. 
Assume that $L> 0$. There exist $N,\epsilon>0$ such that we have $$u_n<L+\epsilon\quad\text{for all } n>N,$$ 
$$-\epsilon<u_{n+1}-u_n(1+u_n)<\epsilon \quad\text{for all }n>N,$$
$$u_m>L-\epsilon \quad \text{for some }m>N.$$
Then, $$u_{m+1}>u_m(1+u_m)-\epsilon>(L-\epsilon)(1+L-\epsilon)-\epsilon=L+L^2+O(\epsilon).$$ So if we choose $\epsilon$ small enough we contradict the fact that $u_{m+1}<L+\epsilon$. Hence $L\le 0$. It is pretty straightforward to show that we cannot have $L<0$ either. So $L=0$.
A: Let $\epsilon \in (0, \frac{1}{4})$ be arbitrary. Then there exists $N = N(\epsilon)$ such that
$$ u_{n+1} \geq u_n^2 + u_n - \epsilon =: f_{\epsilon}(u_n) $$
for all $n \geq N$. Now we investigate some properties of $f_{\epsilon}$. Let $X = [-\tfrac{1}{2}, \infty)$.


*

*(P1) We know that the minimum of $f_{\epsilon}$ is exactly $f_{\epsilon}(-\tfrac{1}{2}) = -\tfrac{1}{4} - \epsilon$, which is greater than $-\tfrac{1}{2}$. This means that $f_{\epsilon} : \Bbb{R} \to X$.

*(P2) $f_{\epsilon}$ is increasing on the interval $X$. Thus if we restrict the domain of $f_{\epsilon}$ onto $X$, then $f_{\epsilon} : X \to X$ is an increasing function.

*(P3) (Fixed points of $f_{\epsilon}$) Notice that $f_{\epsilon}(x) = x$ if and only if $x = \pm \sqrt{\epsilon}$. Also, by a simple argument, we can prove that
$$ \lim_{n\to\infty} f_{\epsilon}^{\circ n} (x) = \begin{cases}
-\sqrt{\epsilon} & \text{if } -\frac{1}{2} \leq x < \sqrt{x} \\
\sqrt{\epsilon} & \text{if } x = \sqrt{\epsilon} \\
+\infty & \text{if } x > \sqrt{\epsilon}.
\end{cases} $$
Let us return to the proof.


*

*By (P1), we have $u_{n+1} \geq f_{\epsilon}(u_n) \geq -\frac{1}{2}$ for all $n \geq N$. In other words, $u_n \in X$ for all $n > N$.

*Lower bound as iterated function. Using (P2) repeatedly, for any $n > N$ and $m \geq 1$ we have
$$ u_{n+m} \geq f_{\epsilon}(u_{n+m-1}) \geq f_{\epsilon}^{\circ 2}(u_{n+m-2}) \geq \cdots \geq f_{\epsilon}^{\circ m}(u_n). \tag{2} $$

*Bounding limsup from above by $\sqrt{\epsilon}$. Assume that we have $u_n > \sqrt{\epsilon}$ for some $n > N$. Then by (2) and (P3),
$$ u_{n+m} \geq f_{\epsilon}^{\circ m}(u_n) \xrightarrow[m\to\infty]{ } +\infty, $$
contradicting the boundedness of $(u_n)$. This implies that $u_n \leq \sqrt{\epsilon}$ for all $n \geq N$.

*Bounding liminf from below by $-\sqrt{\epsilon}$. Again, by (2) and (P3), we have
$$ \liminf_{n\to\infty} u_n = \liminf_{m\to\infty} u_{n+m} \geq \lim_{m\to\infty} f_{\epsilon}^{\circ m}(u_n) \geq -\sqrt{\epsilon}. $$

*Conclusion. By the previous two steps, we know that
$$ -\sqrt{\epsilon} \leq \liminf_{n\to\infty} u_n \leq \limsup_{n\to\infty} u_n \leq \sqrt{\epsilon}. $$
Therefore by taking $\epsilon \downarrow 0$, we conclude that $(u_n)$ converges to $0$.
A: 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 $\lvert u_{n} \rvert$ < 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 $kr^2-2M=A>0$. 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 )
