Recursive sequence with first term defined in interval

Question

I have to prove that the recursive sequence $$a_{n+1}=2a_n(a_n+1)$$ converges for $-1 \le a_0 \lt 0$.

I began by trying to prove that the sequence is decreasing: $$ a_n - a_{n+1}>0, $$ $$ a_n-2a_n(a_n+1)>0, $$ $$ a_n-2a_n^2 - 2a_n>0. $$ I don't know how to prove that inequality since $a_0$ is defined on an interval. What should I do next?

Answer 1

$a_0=-1$ means $a_1=0$ and $a_n=0$ so assume that $-1<a_n<0$ by induction $0< a_n<1$ since, if $-1<a_n<0$ then by adding one $0< a_n+1 < 1$. $a_{n+1}=a_n(a_n+1)$ is the product of $0< a_n+1 < 1$ and $-1< a_n < 0$ so the product will be strictly between -1 and 0. For example $\frac{-1}{2}\times\frac{1}{2}=\frac{-1}{4}<0$ and $\frac{-1}{4}>-1$. Okay now showing it is actually increasing is possible $a_{n+1}=a_n(a_n+1)>a_n$ since $a_n<0$ and $0<a_n+1<1$. Bounded increasing sequences converge.

Answer 2

Consider the function $f(x)=2x^2(x+1)$. Basic analysis allows you to show that this function sends $[-1,0]$ onto $[-1/2,0]$, so that if $a_0 \in [-1,0]$, then $a_n$ also for all $n$.

We now seperate the values in $[-1,0)$ into $3$ cases : either $a_0<-1/2$, either $a_0=-1/2$, either $a_0>-1/2$.

The second case is easy as $-1/2$ is fixed under the function $f$. In the first case, we notice that if $a_0=-1$, the sequence gets stuck on $0$ and stays there. If $-1<a_0<-1/2$, then $a_1 >-1/2$ and we are in the third case. We now only need to show that if $a_0>-1/2$ then the sequence converges. This is simple, for you can show that for all $x \in (-1/2,0)$, $f(x)<x$. The sequence is thus decreasing and bounded, and so converges.

You can also show that it always converge to $-1/2$ except when $a_0=-1$. Give it a try!

Answer 3

Hint 1: Try a few cases in Excel and see where $a_n$ goes.

Hint 2: complete the square on your RHS.

Next $$a_{n+1} + \frac 12 = 2a_n^2+2a_n+\frac 12 = 2\left(a_n+\frac 12\right)^2.$$ Can you end it now?

Next: $$2a_{n+1}+1 = (2a_n+1)^2 \implies 2a_n+1 = \left(2a_0+1\right)^{2^n} \to 0, \text{ if } a_0 > -1;\\ 2a_n+1 \equiv 1, \forall n>0, \text { if } a_0=-1.$$

Answer 4

If $a_{n+1}=2a_n(a_n+1)$ then $2a_{n+1} =4a_n(a_n+1) =(2a_n+1)^2-1 $ so $2a_{n+1}+1 =(2a_n+1)^2 $.

Let $b_n = 2a_n+1 $. Then $b_{n+1} =b_n^2 $ so $b_{n+k} =b_n^{2^k} $.

Since $-1 \le a_0 < 0$, $-1 \le b_n < 1$.

If $a_0 = -1, a_n = 0$ for $n \ge 1$.

If $a_0 >-1$, then $-1 < b_0 < 1$ so $b_n =b_0^{2^n} \to 0$ so $a_n =\dfrac{b_n-1}{2} \to -\dfrac12 $.