TITLE: $u_{n+1}-u_n-u_n^2\to 0$ implies $u_n$ goes either to $0$ or to $+\infty$
QUESTION [7 upvotes]: Let $u_n$ be a real sequence such that $\displaystyle u_{n+1}-u_n-u_n^2\to_{\infty} 0$.
Prove that either $u_n\to 0$ or $u_n \to +\infty$

Progress

If $u_n$ is bounded,

it has a convergent subsequence $u_{n_k}$ that goes to $\beta$.
By assumption, $\displaystyle u_{n_k+1}\to \beta^2+\beta$
This proves that if $\beta$ is an accumulation point of $u_n$, then $\beta^2+\beta$ also is.
This forces $\beta \in (-2,0]$ (otherwise the sequence is not bounded)
And in this case, iterating and using the closedness of the set of accumulation points yields that $0$ is an accumulation point of $u_n$.
I should prove next that $\beta=0$, but I can't.
EDIT: the crucial point that I missed is going backward (rewriting $u_n-u_{n-1}-u_{n-1}^2\to 0$), as Krokop did in his answer.

If $u_n$ is unbounded,

$u_n$ has a subsequence that goes to $+|-\infty$.
EDIT: this part still lacks a slick and elegant proof

REPLY [4 votes]: Let $E$ be the set of  accumulation point of $(u_n)$. You have already prove that  $\beta^2+\beta\in E$ 
$u_n$ is bounded, one can extract a convergent subsequence of the sequence $u_{\phi(n)-1}$ and its limit $\mu$ verifies $\mu^2+\mu=\beta$. So $E$ is invariant under $f$. Where $f:\mathbb{R}\rightarrow \mathbb{R}$, $x\mapsto x^2+x$.
Let $m=\inf E \quad\text{and}\quad M=\sup E$. We have $f(M)\le M$ then $M=0$.
Moreover,
If $\lambda\in E$ then there exist $\alpha\in E$ such that $f(\alpha) = \alpha^2+\alpha = \lambda$ : Every element of $E$ is in the image of $f$, then $m\ge\frac{-1}{4}$, and $[m^2+m,M^2+M] = f(E) = E = [m,M]$ because $f$ is increasing on $[\frac{-1}{4},0]$. So $m^2 + m = m$ and $M^2 + M = M$. So $m = M = 0$.
Therefore $E=\{0\}$ and your result follow.