Limits of a recursively defined sequence

Question

Let $x_1=a$ and define a sequence $\left(x_n\right)$ recursively by: $x_{n+1} = \dfrac{x_n}{1 + \frac{x_n}{2}}$

For what values of $a$ is it true that $x_n$ approaches $0$?

Answer 1

Whenever the sequence converges (to $L$, say) we must have $L=\frac L{1+\frac L2}$, i.e. $L=0$. So the question is just: For which $a$ does the sequence converge?

If $a>0$ then by induction all $x_n>0$ so that the sequence is strictly decreasing and bounded from below, hence convergent.

If $a=0$ the sequence is constant, hence also convergent.

If $a<-2$ then $x_2>0$ and the sequence converges as in the case $a=x_2>0$.

If $a=-2$ the recursion fails to define $x_2$, i.e. we don ont even have a sequence.

If $-2<a<0$, the sequence starts decreasing as long as $-2<x_n<0$. It cannot remain bounded from below by $-2$ as otherwise it would converge to some number on $[-2,0)$, which is impossible. Thus after finitely many steps we obtain some $x_n<-2$ and from then on convergence as shown above. Or we hit $-2$ exactly and the recursion fails. To see for which $a$ we hit $-2$, solve the recursion for $x_n$: $$ x_n = \frac{2x_{n+1}}{1-x_{n+1}}.$$ Consider the sequence $y_n$ given by $y_1=2$ and $y_{n+1}=\frac{2y_n}{1+y_n}$. If and only if $-a=y_n$ for some $n$, our original sequence will hit $-2$ (for we see that $x_k=-y_{n+1-k}$ for $1\le k\le n$). One readily shows that $y_n=\frac{2^n}{2^n-1}$.

Summary: If $a=-\frac{2^n}{2^n-1}$ for some $n$, the sequence is only finite. In all other cases it converges to $0$.