$u_{n+1}-u_n-u_n^2\to 0$ implies $u_n$ goes either to $0$ or to $+\infty$ 
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
 A: 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.
A: The statement to be proved is equivalent to saying that if $u_n$ does not converge to $\infty$, then for all $\epsilon > 0$ there is an $N$ such that for $n > N$ you have
$$|u_n| < \epsilon \tag 1$$
Furthermore, the condition that $\lim_{n \rightarrow \infty} u_{n+1} - u_n - u_n^2 = 0$ has the consequence that for all $\eta > 0$ there is an $M$ such that if $n > M$ you have
$$u_{n+1} > u_n + u_n^2 - {\epsilon^2 \over 2} \tag 2$$
In particular $(2)$ implies
$$u_{n+1} - u_n > - {\epsilon^2 \over 2} \tag 3$$
Here is the proof that $(1)$ holds:
Suppose $n > N,M$ and $u_n > \epsilon$. Then $(2)$ says that $u_{n+1} > u_n +{\epsilon^2 \over 2}$. This continues with each iteration and the sequence goes to infinity, which we are supposing does not happen.
Suppose now $n > N,M$ with $u_n <-\epsilon$. Then again $u_{n+1} > u_n + {\epsilon^2 \over 2}$ and the sequence will start to increase. By $(2)$, it can only decrease after that if $u_k^2 < {\epsilon^2 \over 2}$, in other words if $|u_k| < {\epsilon \over \sqrt{2}}$. But by $(3)$ the amount it can decrease is at most $\epsilon^2 \over 2$, and so you will never be below $-\epsilon$ again; you can only go as far as ${\epsilon \over \sqrt{2}} - {\epsilon^2 \over 2}$ before the sequence starts going up again. Hence eventually $u_k$ stays above $-\epsilon$. 
A: Doing unbounded case for now.
First we prove that $u_n$ can't be bounded above. If $u_n$ is bounded below then we are done so suppose $u_n$ isn't. Fix $\epsilon >0$ and assume $|u_{n+1}-u_n-u_n^2 |< \epsilon$ for all $n >N$ for some $N$. For all $C>2$ there exist $m \in \mathbb{Z}_{>0}$ s.t $u_M < -C$ and we can assume $m> N$. Then $|u_{m+1}-u_m -u_m^2| < \epsilon$. Choosing $C$ large enough and $\epsilon$ small enough we can gather $u_{m+1}>C$.
In fact with this construction we prove that $(u_n)$ is eventually increasing sequence which is not bounded above so it must tend to infinity. 
I think similar epsilon argument should work for the first case as well but I haven't tried yet.
A: I will work by absud:
Suppose that $u_n\not\to0$ and $u_n\not\to+\infty$. So either $u_n\rightarrow\ell\in\mathbb{R}\backslash\{0\}$ or $u_n\rightarrow-\infty$.


*

*Suppose that $u_n\rightarrow\ell\in\mathbb{R}\backslash\{0\}$. Since $u_{n+1}-u_n-u_n^2\rightarrow0$ then $\ell-\ell-\ell^2=0$. 
Thus, $\ell=0$. 
Absurd. 
Therefore, if $u_n$ converges it must converges to $0$. 

*If $u_n$ diverges, $u_n\rightarrow-\infty$, then we have 
$$\lim_{n\to+\infty}u_{n+1}-u_n-u_n^2=\underbrace{u_{n+1}}_{\to-\infty}\;\;\underbrace{\underbrace{-\;u_n}_{\to+\infty}\;\;\underbrace{(u_n+1)}_{\to-\infty}}_{-\infty}=-\infty.$$
Absurd.
Therefore, if $u_n$ diverges it must diverges to $+\infty$ . 
