Convergence of a Recursively Defined Series: $a_n =5$, $a_{n+1} = 1/(6-a_n)$?

Question

Determine whether the series converges or diverges. I have no idea how to go about this when it isn't defined in terms of n. Help is appreciated. $$a_n =5, a_{n+1} = 1/(6-a_n)$$

Answer 1

One can prove by induction that $a_n <1$ for all $n>2$.

Induction basis: $a_3=\frac {1}{5} <1$

Induction step: $$a_{n+1}=\frac {1}{6-a_n}<\frac {1}{6-1}<\frac {1}{5}<1$$ We used the induction hypothesis for the first inequality sign.

One can then prove that the sequence is strictly decreasing: $$a_{n+1}=\frac {1}{6-a_n}<\frac {1}{6a_n-a_n}=\frac {1}{5}a_n<a_n$$ We used our statement we proved before for the first inequality sign.

And since we proved that $a_n<1$ for $n>2$ it follows that $6-a_n >0$ and therefore all the elements of the sequence are positive.

All in all, we have proven that the sequence is strictly decreasing and bounded from below which implies its convergence.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Write $\ds{a_{n}}$ as $\ds{x_{n} \over y_{n}}$ with $\ds{x_{n + 1} = y_{n}}$ and $\ds{y_{n + 1} = -x_{n} + 6y_{n}}$. $\ds{x_{n} \over y_{n}}$ satisfies the recurrence and \begin{align} &{x_{n + 1} \choose y_{n + 1}}\ =\ \overbrace{\pars{\begin{array}{rr}0 & 1 \\ -1 & 6\end{array}}} ^{\ds{\mathsf{M}}}\ {x_{n} \choose y_{n}} = \pars{\begin{array}{rr}0 & 1 \\ -1 & 6\end{array}}^{n}{x_{1} \choose y_{1}} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, \lambda^{n}\,\mathbf{u}\mathbf{u}^{T} {x_{1} \choose y_{1}} \propto \lambda^{n}\,\mathbf{u} \\[5mm] &\ \mbox{with}\quad \left\{\begin{array}{l} \ds{x_{1} = 5} \\ \ds{y_{1} = 1} \end{array}\right. \end{align} where $\ds{\lambda}$ is the largest, in magnitude, eigenvalue of $\ds{\mathsf{M}}$ and $\ds{\mathbf{u} \propto {3 - 2\root{2} \choose 1}}$ the correspondent eigenvector. Note that $\ds{0 \not= \mathbf{u}^{T} {x_{1} \choose y_{1}} \propto 16 - 10\root{2}}$.

$$ \mbox{So,}\qquad {x_{n + 1} \over y_{n + 1}} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\, {3 - 2\root{2} \over 1} = \bbx{\ds{3 - 2\root{2}}} $$

Answer 3

If $(a_n)_n$ converges to a limit $L$ then $L=1/(6-L),$ equivalently $L^2-6L+1=0,$ so $L=3\pm 2\sqrt 2.$ Experimentally, the values $a_1=5,$ $a_2=1,$ $a_3=1/5,$ $a_4=5/29$ suggest that $(a_n)_n$ appears to be converging to the smaller value $3-2\sqrt 2.$

So let $A=3-2\sqrt 2$ . Let $a_n=A+d_n.$

We have $1-6A+A^2=0$ and $|A|=A<1.$

Suppose $|d_n|<1$ for some $n$. Then $|6-a_n|=6-(A+d_n)|>4.$ And also $$|d_{n+1}|=|a_{n+1}-A|=\left|\frac {1}{6-a_n}-A\right|=\left|\frac {1}{6-(A+d_n)}-A\right|=$$ $$=\frac {|1-6A+A^2-Ad_n|}{|6-(A+d_n)|}|=\frac {|Ad_n|}{|6-(A+d_n)|}\leq|d_n|/4.$$ Therefore by induction on $m\geq n$ we have $|d_m|<1$ for all $m\geq n .$

But then,also by induction on $m\geq n ,$ we have $|d_{m+1}|\leq |d_m|/4$ for all $m\geq n.$

So $\lim_{m\to \infty}d_m=0$ because $|d_m|\leq 4^{-(m-n)}|d_n|$ for all $m\geq n.$ And $d_m\to 0$ is equivalent to $a_m\to A.$