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We define the sequence $(u_n)_{n=1}^\infty$by:$$u_{n+1}=1+\frac{1}{u_n}$$ How can I find the limit of this sequence as it goes to infinity?
By induction, I can prove that it is bounded above and below. I have also proved that $$u_{n+2}-u_n=\frac {u_n-u_{n-2}}{(1+u_n)(1+u_{n-2})}$$ Therefore, I can show that $\lim_{n\to\infty} u_{2n}$ and $\lim_{n\to\infty} u_{2n+1}$ exist. However, I am unable to find the limits themselves or a an explicit formula. How does one go around doing this? Are there any standard methods?

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    $\begingroup$ $u_\infty=1+\dfrac 1{u_\infty}$. This is a quadratic equation but only the one which falls in the bound you have proved is the real limit. $\endgroup$ Commented Jun 12, 2016 at 17:01
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    $\begingroup$ Suppose $\lim u_n=L$. What is $\lim u_{n+1}$? $\endgroup$ Commented Jun 12, 2016 at 17:01
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    $\begingroup$ At convergence, there is no difference anymore between $u_{n+1}$ and $u_n$. What can you conclude ? $\endgroup$
    – user65203
    Commented Jun 12, 2016 at 17:02
  • $\begingroup$ I had never thought of that, it is such a beautiful idea $\endgroup$
    – GuPe
    Commented Jun 12, 2016 at 18:59

3 Answers 3

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The candidate limit value $\ell$, satisfying $$ \ell=1+\frac{1}{\ell}, $$ with solutions $\ell=\frac{1}{2}(1\pm\sqrt{5}) $, suggests a relation with the Fibonacci sequence $\{F_n\}$. In fact, if you put $$ u_n=\frac{F_{n+1}}{F_n}, $$ you have the recurrence relation $$ u_{n+1}=\frac{F_{n+2}}{F_{n+1}}=\frac{F_{n+1}+F_n}{F_{n+1}}=1+\frac{1}{u_n}. $$ Therefore, you can use all the information about the Fibonacci sequence. It is not necessary to repeat that here.

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  • $\begingroup$ I really like how you offered an additional relationship with other well known sequences. $\endgroup$ Commented Jun 12, 2016 at 22:44
  • $\begingroup$ The OP did not show that $u_n$ has a limit, but by relating $u_{k+2}$ to $u_k$, a similar argument shows that $u_{2n}$ and $u_{2n+1}$ both converge to a solution of the same equation. $\endgroup$
    – user14972
    Commented Jun 13, 2016 at 0:30
  • $\begingroup$ Very nice (+1). If you had not guessed the relationship with the Fibonacci sequence from the limit, how would you have derived it using only the recurrence relation provided? $\endgroup$ Commented Jun 13, 2016 at 17:09
  • $\begingroup$ Well, if I hadn't noted the appearance of the golden ratio, I think I wouldn't have thought to the Fibonacci numbers at all. Unfortunately I cannot provide any general advice. Maybe I can only make a remark about some vague similarity with the differential equations case. Here, from a linear second order (i.e involving $n$, $n+1$, $n+2$) recurrence relation (the one of the Fibonacci numbers), we can obtain a non-linear first order recurrence relation. Similarly, from a second order linear differential equations, one can obtain an equivalent first order non linear equation: the Riccati equation $\endgroup$
    – guestDiego
    Commented Jun 13, 2016 at 17:55
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the initial term $u_0$ can not be $-1$ or $0$, so if the limit of $ u_n $ exist, then it is necessarily the limit $l$ is a positif root of the $l^2-l-1$, following it is $\frac{1+\sqrt{5}}{2}$.

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  • $\begingroup$ What about $u_0 = (1 - \sqrt{5})/2$? $\endgroup$
    – user14972
    Commented Jun 13, 2016 at 0:32
  • $\begingroup$ the sequence is stationair $u_n=u_0$ $\endgroup$
    – m.idaya
    Commented Jun 14, 2016 at 9:58
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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{u_{n + 1} = 1 + {1 \over u_{n}} = {u_{n} + 1 \over u_{n}}}\,,\qquad u_{1} \not= 0\tag{1}$.

Let's consider the pair of sequences $\ds{p_{n + 1} = p_{n} + q_{n}}$ and $\ds{q_{n + 1} = p_{n}}$. Note que $\ds{p_{n}/q_{n}}$ satisfies $\pars{1}$. With $\ds{r_{n} \equiv {p_{n} \choose q_{n}}}$ we note that $\ds{r_{n + 1} = \mathsf{A}r_{n}}$ such that $\ds{r_{n} = \mathsf{A}^{n - 1}r_{1}}$ where

\begin{align} \mathsf{A} & = \pars{\begin{array}{cc}\ds{1} & \ds{1} \\ \ds{1} & \ds{0}\end{array}} = \half\pars{\begin{array}{cc}\ds{1} & \ds{0} \\ \ds{0} & \ds{1}\end{array}} + \pars{\begin{array}{cc}\ds{0} & \ds{1} \\ \ds{1} & \ds{0}\end{array}} + \half\pars{\begin{array}{cc}\ds{1} & \ds{0} \\ \ds{0} & \ds{-1}\end{array}} = \half\,\sigma_{0} + \vec{a}\cdot\vec{\sigma} \\[3mm] & \sigma_{0}\quad \mbox{is the}\ identity\ matrix\ \mbox{and}\quad \vec{a} \equiv \hat{x} + \half\,\hat{z} \end{align}

$\ds{\vec{\sigma}}$ is a Pauli Matrix Vector. Then, $\ds{r_{n} = \pars{\half\,\sigma_{0} + \vec{a}\cdot\vec{\sigma}}^{n - 1}r_{1}}$. Note that $\ds{\vec{a}\cdot\vec{\sigma}} = \pars{\begin{array}{cc}\ds{1/2} & \ds{1} \\ \ds{1} & \ds{-1/2}\end{array}}$ and $\ds{\pars{\vec{a}\cdot\vec{\sigma}}^{2} = {5 \over 4}\,\sigma_{0}}$.


With the vectors \begin{equation} s_{-} \equiv {0 \choose 1}\quad\mbox{and}\quad s_{+} \equiv {1 \choose 0}\,, \qquad u_{n} = {s_{+}^{\mathrm{T}}\pars{\sigma_{0}/2 + \vec{a}\cdot\vec{\sigma}}^{n - 1}r_{1} \over s_{-}^{\mathrm{T}}\pars{\sigma_{0}/2 + \vec{a}\cdot\vec{\sigma}}^{n - 1}r_{1}} \tag{2} \end{equation}
We have to evaluate $\ds{\pars{\sigma_{0}/2 + \vec{a}\cdot\vec{\sigma}}^{n - 1}}$. For this purpose, lets consider $\ds{\expo{\lambda\pars{\sigma_{0}/2 + \vec{a}\cdot\vec{\sigma}}} = \expo{\lambda/2}\expo{\lambda\vec{a}\cdot\vec{\sigma}}}$. Note that $\ds{\expo{\lambda\vec{a}\cdot\vec{\sigma}}}$ satisfies

\begin{align} &\pars{\totald[2]{}{\lambda} - {5 \over 4}}\expo{\lambda\vec{a}\cdot\vec{\sigma}} = 0 \quad\mbox{with}\quad \left.\expo{\lambda\vec{a}\cdot\vec{\sigma}}\right\vert_{\ \lambda\ =\ 0} = \sigma_{0}\,,\quad \left.\totald{\expo{\lambda\vec{a}\cdot\vec{\sigma}}}{\lambda}\right\vert_{\ \lambda\ =\ 0} = \vec{a}\cdot\vec{\sigma} \\[3mm] \imp\quad &\ \expo{\lambda\pars{\sigma_{0}/2 + \vec{a}\cdot\vec{\sigma}}} = \expo{\lambda/2}\cosh\pars{{\root{5} \over 2}\,\lambda}\sigma_{0} + {2 \over \root{5}}\, \expo{\lambda/2}\sinh\pars{{\root{5} \over 2}\,\lambda}\vec{a}\cdot\vec{\sigma} \\[3mm] = &\ \half\pars{\sigma_{0} + {2 \over \root{5}}\,\vec{a}\cdot\vec{\sigma}}\exp\pars{\varphi\lambda} + \half\pars{\sigma_{0} - {2 \over \root{5}}\,\vec{a}\cdot\vec{\sigma}}\exp\pars{\Phi\lambda} \end{align}

where $\ds{\varphi \equiv {1 + \root{5} \over 2}}$ and $\ds{\Phi \equiv {1 \over \varphi} = {1 - \root{5} \over 2}}$ are the Golden Ratio and the Conjugated Golden Ratio, respectively.

Then, \begin{align} \pars{\half\,\sigma_{0} + \vec{a}\cdot\vec{\sigma}}^{n - 1} = \half\pars{\sigma_{0} + {2 \over \root{5}}\,\vec{a}\cdot\vec{\sigma}} \varphi^{n - 1} + \half\pars{\sigma_{0} - {2 \over \root{5}}\,\vec{a}\cdot\vec{\sigma}} \Phi^{n - 1} \end{align}

With this expression and $\pars{2}$ we can get an $\underline{\mbox{explicit formula for}\ u_{n}}$ which is a cumbersome algebraic task.


However, the limit $\ds{n \to \infty}$ is somehow straightforward since $\ds{\varphi > \Phi > 1}$. In addition $$ \sigma_{0} + {2 \over \root{5}}\,\vec{a}\cdot\vec{\sigma} = \pars{% \begin{array}{cc} \ds{1 + {1 \over \root{5}}} & \ds{2 \over \root{5}} \\ \ds{2 \over \root{5}} & \ds{1 - {1 \over \root{5}}} \end{array}} $$

With expression $\pars{2}$ $\ds{\pars{~\mbox{note that}\ u_{1} \not= 0~}}$: \begin{align} \color{#f00}{\lim_{n \to \infty}u_{n}} & = {\pars{1 \quad 0}\pars{\sigma_{0} + 2\,\vec{a}\cdot\vec{\sigma}/\root{5}} \ds{{u_{1} \choose 1}} \over \pars{0 \quad 1}\pars{\sigma_{0} + 2\,\vec{a}\cdot\vec{\sigma}/\root{5}} \ds{{u_{1} \choose 1}}} = {2/\root{5} + \pars{1 + 1/\root{5}}u_{1} \over 1 - 1/\root{5} + 2u_{1}/\root{5}} \\[3mm] & = \color{#f00}{{1 + \root{5} \over 2}} = \color{#f00}{\varphi}\,,\quad\forall\ u_{1} \not= 0 \end{align}

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