Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

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 $0$).

share|improve this answer

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?

share|improve this answer
    
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 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 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 at 7:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.