I was exploring a Fibonacci-like sequence,
$$\frac{1}{u_{n+1}} = \frac{1}{u_n} + \frac{1}{u_{n-1}}$$
Solving it directly, $u_n=(A(\frac{1+\sqrt5}{2})^n+B(\frac{1-\sqrt5}{2})^n)^{-1}$ which converges to $0$ whenever it is defined. But I am looking for a more elementary approach to reach this same conclusion.
Firstly, assuming that it converges to say $L$, then $L=\frac{1}{1/L+1/L}$, thus $L=0$ since $\lim_{L\to0}L-\frac{1}{2/L}=0$
Now, if $u_0$ and $u_1$ having the same sign, consider $\frac{1}{u_{n+1}} – \frac{1}{u_n}=\frac{1}{u_{n-1}}$
If $u_0,u_1>0$, then $u_n>0$ by induction, so $\frac{1}{u_{n+1}} – \frac{1}{u_n}>0 \implies u_{n+1}<u_n$; thus $u_n$ converges to $0$ (I think by monotone convergence theorem?). Likewise for $u_0,u_1<0$.
However, if they have different signs, such a method does not work. From graphing a few initial conditions, the sequence will alternate a few times before settling down and converge $0$. I try to prove that there exists $N>0$ such that $u_N$ and $u_{N+1}$ have the same sign, thus it will converge by earlier result. But I'm not sure what approach to use to solve this.
What I tried thus far was to consider $Q_n = u_n – u_{n-1}$ and show that $Q_n$ is decreasing, which isn't true, I also tried to consider the function $\frac{1}{1/(x-a)+1/(x-b)}$ where $a,b>0$. Hopefully there will be some properties I can make use of but to no avail too.
Any guide/direction on how to prove convergence for the case the initial conditions have different signs are much appreciated. Thank you!
background: I am learning recurrence relation at a highschool level. We have learnt how to check for convergence, increasing/decreasing sequence through algebraic and graphical means, and finding closed forms.
Best Answer
The $u$ sequence does not always converge to $0$. In particular, when $u_1=-\phi u_0$, it does not converge.
Let us consider the reciprocal sequence: $$r_n=\frac{1}{u_n}$$ $$r_n=r_{n-1}+r_{n-2}$$
Which follows the fibonacci recurrence. Thus, $r_n=a(\phi)^n+b(-\phi)^{-n}$ for some $a$ and $b$. Notice that $(u_n \rightarrow 0) \iff (r_n \rightarrow \infty) \iff (a \neq 0)$.
Thus if we take $a=0$ and $b=1$ (equivalently take $r_0=1$ and $r_1=(-\phi)^{-1})$) then $r_n=(-\phi)^{-n} \rightarrow 0$. Therefore, the corresponding $u$ sequence does not converge.