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Define the recursive sequence $(a_n)$ as $a_{n+1} = a_n^2 + a_n$ where $n \in \mathbb{N}$. We want to find $a_1 \in \mathbb{R}$ such that $(a_n)$ converges to a limit that depends on $a_1$.

Attempt: Let $a_1 = k$, then $a_2 = k^2 + k$, $a_3 = (k^2 + k)^2 + (k^2 + k)$, ... clearly if we have |k| > 1 then the sequence $(a_n)$ will diverge. From this we deduce the interval of convergence must be when $|k| \le 1$. Suppose $(a_n) \to L$ then $a_{n+1} = a_n^2 + a_n$ $\implies L = L^2 + L \implies L = 0$. Surely if $k = 0$ this will occur as $a_n = 0$ for all n. But it is not clear to me if there are other $k$ such that the recursive sequence converges.

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up vote 2 down vote accepted

Since $x_{n+1}-x_n=x_n^2\geq0$, your sequence $(x_n)$ is always increasing, so either it converges to a real number, or it diverges to $+\infty$.

Suppose that the sequence converges, and let $L\in\mathbb R$ be the limit. Taking $n\to\infty$ in the equality $x_{n+1}=x_n^2+x_n$ we get $L=L^2+L$, so necessarily $L=0$ and $x_n\leq0$ for all $n$ (recall that the sequence is increasing). On the other hand, if $f(x)=x^2+x$ then $f(x)>0$ for $x<-1$ and $-\frac14\leq f(x)\leq0$ for $-1\leq x\leq0$. This implies that necessarily we have $x_1\in[-1,0]$. Conversely, if $-1\leq x_1\leq 0$, then you can prove by induction on $n$ that $x_n\in[-1,0]$, so your sequence is increasing and bounded above, so it converges (to $0$).

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Try $-1\leq k\leq 0$. For instance $x_1=-1/2$, $x_2= -1/4$, $x_3=-3/16$, etcetera. A sequence that's monotonically increasing and bounded above must be convergent, right?

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Ohh, so in this case we need to prove that if $k \in [-1,0]$ then $(x_n)$ is always monotonically increasing bounded sequence, with the upper bound being 0, then we would be done. – user77404 Feb 3 '14 at 5:20
If $k \in (0,1)$ wont the sequence converge to a limit of 1? When $k \in (-1,0)$ then the sequence converge to a limit of 0? – user77404 Feb 3 '14 at 6:26
It is easy to verify that if there is a limit, that limit must be $0$. If at some stage we have $x_n\gt 0$, then since $x_{n+1}\gt x_n$, the limit cannot be $0$. Thus it is enough to consider $x_1\le 0$. If $x_1\lt -1$, then $x_2\gt 0$, so convergence fails. Thus the only candidates are $-1\le x_1\le 0$. As in the answer by JPi, in that case we have convergence. – André Nicolas Feb 3 '14 at 7:11

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