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I have the following recurrence relation

\begin{align} u_{0} &= 1 \nonumber \\ u_{1} &= 2 \nonumber \\ u_{n} &= u_{n-2} + u_{n-1} +1 \end{align}

Now I want to prove that the ratio after $n$ (where $n$ is large) remains constant. I can see this by simple putting the equation in excel and then see that the ratio grows to the Golden Ratio (i.e. $\frac{1+\sqrt{5}}{2}$).

Hence I want to prove $$\lim_{n \rightarrow \infty} \frac{u_{n+1}}{u_{n}} = C,$$ for some $C \in \mathbb{R}$ (more specifically $C = \frac{1+\sqrt{5}}{2}$.

I find $$\frac{u_{n+1}}{u_{n}} = \frac{u_{n-2}+u_{n-1}+1}{u_{n-1}+u_{n}+1},$$ but this does not really help me. I would say we can ommit the $+1$ in the equations since $u_{n}$ grows large but this is not really mathematically sound.

Any suggestions are more than welcome.

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2 Answers 2

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Define the sequence $(v_n)_{n \in \mathbb{N}}$ by $v_n:=u_n+1$ so that $v_0=2, v_1=3, v_n = v_{n-1}+v_{n-2}$ for all $n \geq 2$. Then an easy induction argument yields $v_n = F_{n+3}$ where $F_n$ denotes the $n$-th term of the Fibonacci sequence (defined by $F_0:=0, F_1:=1$ and $F_n = F_{n-1}+F_{n-2} \forall n \geq 2$), whence for all $n \geq 0$ we have $u_n = F_{n+3}-1 = \frac{1}{\sqrt{5}} (\alpha^{n+3} - \beta^{n+3})$ where $\alpha = \frac{1+\sqrt{5}}{2}, \beta = \frac{1-\sqrt{5}}{2}$, using the well-known Binet's Formula (which can be derived simply by looking at the characteristic equation of the Fibonacci sequence). This finally yields $$\lim_{n \rightarrow \infty} \frac{u_{n+1}}{u_n} = \lim_{n \rightarrow \infty} \frac{F_{n+4}-1}{F_{n+3}-1} = \lim_{n \rightarrow \infty} \frac{F_{n+1}-1}{F_n-1} = \lim_{n \rightarrow \infty} \frac{\alpha^{n+1} - \beta^{n+1} - \sqrt{5}}{\alpha^n - \beta^n - \sqrt{5}} = \alpha \left( \lim_{n \rightarrow \infty} \frac{1 - \lambda^{n+1} - \frac{\sqrt{5}}{\alpha^{n+1}}}{1 - \lambda^n - \frac{\sqrt{5}}{\alpha^n}} \right) = \alpha = \frac{1+\sqrt{5}}{2}$$ where we have used $\beta<1<\alpha \implies \lambda := \frac{\beta}{\alpha}<1 \implies \lim_{n \rightarrow \infty} \lambda^n=0$.

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Note that we have $\frac{u_{n+1}}{u_n}<2 $ and the sequence is monotonically increasing, and bounded above, it guarantees the existence of the limit.

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  • $\begingroup$ How do you know the fraction is smaller than $2$? $\endgroup$
    – user735323
    Commented Jan 25, 2020 at 22:43
  • $\begingroup$ See that $u_{n+1}=u_{n-1}+u_n<u_n+u_n$, fibonacci sequence is an increasing sequence. $\endgroup$
    – aud098
    Commented Jan 26, 2020 at 5:01
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    $\begingroup$ This is not the Fibonacci Sequence, but it is clear that $u_n>0$ for all $n$ (induction), and so the recursion itself yields that the sequence is increasing, and now aud098's argument works. $\endgroup$
    – asrxiiviii
    Commented Jan 26, 2020 at 7:58

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