3

Question: Let $a_n$ be a sequence given by a starting seed $s$, with $$ a_0 = s$$ an $$ a_n = 1 + \frac{1}{a_{n-1}} $$ Does $a_n$ converge for every $s \in \mathbb{R}$?

This question comes from the fraction tower $1+\frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}$. By choosing a starting seed we may discuss the behavior (convergence) of the sequence $a_n$. Clearly if the sequence converges, it must be either $\phi$ or $\frac{-1}{\phi}$. Using ratio test $$\left|\frac{a_n - a_{n-1}}{a_{n-1} - a_{n-2}}\right| = \left|\frac{-1}{a_{n-2}+1}\right|$$ I found out that $\phi$ is the only possible solution except when $s=\frac{-1}{\phi}$(in which case would imply that $a_{n-1}-a_{n-2}$ would be $0$ for some $n$, and imply that the sequence is precisely $a_n = \frac{-1}{\phi}$.). But I have yet to prove that any starting seed $s \in \mathbb{R}$ except $\frac{-1}{\phi}$ would induce a convergent sequence (but I found some posts on MSE only discuss the convergence of the case where $s=1$ using the monotonicity).

How can I prove the statement above? (or is it correct?) Thanks in advance.

Some $s$ would make some term of the sequence undefined, namely $0, -1, -\frac{1}{2}, ...$, so one should exclude these cases.

Angelo
  • 12,328
Lab
  • 573
  • 2
  • 16

1 Answers1

1

I solved this problem by discussing case by case:

Case 1: $s>0$. This can be done by discussing the monotonicity of the odd and even terms of $a_n$.

Case 2: It can be seen that $(-\infty, -1)$ will map to a positive number in 1 time. $(-1/2, 0)$ can be map to positive number in 2 times, $(-1, \frac{-2}{3})$ will map to positive in 3, etc. These numbers (namely $0, \frac{-1}{2}, \frac{-2}{3}, \frac{-3}{5}, ...$ given by the (negative) quotient of adjacent terms of fibonacci sequence $F_n$) converges to $\frac{-1}{\phi}$ and thus let

$$ u_n = -\frac{F_{n}}{F_{n+1}} = \{-1, \frac{-1}{2}, \frac{-2}{3}, \frac{-3}{5}, ...\}$$ then let $I_0 = (u_2, 0), I_1= (u_1, u_3), I_2 = (u_4, u_2), I_4 = (u_3, u_5)$ etc we have $ \bigcup I_n = (-1, 0) \setminus \{u_n\}$ and hence choosing any points except points in $u_n$ will eventually make $a_n$ converge(and as for $s=u_n$, $a_n$ would be undefined for some term).

Lab
  • 573
  • 2
  • 16