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Consider the recursive sequence \begin{equation*} \begin{split} a_{n+1} = \frac{5}{6-a_n} \quad \textit{with} \quad a_1 = 4. \end{split} \end{equation*} Prove that the sequence $(a_n)$ converges and find its limit, by working out the following steps.
1. First assume that the limit $L = \lim_{n\to \infty} a_n$ exists and find its possible values.
Let $L = \lim_{n\to\infty}a_n$. Then we get \begin{equation*} \begin{split} L &= \lim_{n\to\infty}a_{n+1} \\ &= \lim_{n\to\infty} \frac{5}{6-a_n} \\ &= \frac{5}{6-L}. \end{split} \end{equation*} So we have $L^2-6L+5 = 0 \Longleftrightarrow (L-1)(L-5) = 0$. So the possible values of $L$ are $L = 1$ or $L = 5$.
2. Starting with the initial value $a_1 = 4$, write down the first five entries in the sequence $(a_n)$. Can you see any pattern?
We have \begin{equation*} \begin{split} a_2 &= \frac{5}{6-a_1} = \frac{5}{6-4} = \frac{5}{2} = 2.5 \\ a_3 &= \frac{5}{6-a_2} = \frac{5}{6-5/2} = \frac{10}{7} \approx 1.42857 \\ a_4 &= \frac{5}{6-a_3} = \frac{5}{6-10/7} = \frac{35}{32} = 1.09375 \\ a_5 &= \frac{5}{6-a_4} = \frac{5}{6-35/32} = \frac{160}{157} \approx 1.01091 \\ a_6 &= \frac{5}{6-a_5} = \frac{5}{6-160/157} = \frac{785}{782} \approx 1.00384. \end{split} \end{equation*}
3. Find the real valued function $f(x)$ defining the sequence, i.e. $a_{n+1} = f(a_n)$.
This is the question I'm having trouble with. Please help!

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  • $\begingroup$ Show that $5>a_n>1\implies 5>a_n>a_{n+1}>1$. So if $5>a_1>1$ then the sequence $(a_n)_{n\in \Bbb N}$ is decreasing, but bounded below by $1,$ so it is a convergent sequence. $\endgroup$ Sep 28, 2019 at 6:22

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Hints. For $3.$ $f(x)=\frac{5}{6-x}$ with $f'(x)=\frac{5}{(6-x)^2}>0$, i.e. the function is ascending. Now $a_2=f(a_1)=\frac{5}{2}<4=a_1$. Then $f(a_2)\leq f(a_1) \Rightarrow a_3\leq a_2$ and by induction $$a_n\leq a_{n-1} \Rightarrow f(a_n)\leq f(a_{n-1}) \Rightarrow a_{n+1}\leq a_n$$ or the sequence is descending.

Let's show that it is also bounded. From $$x\in[1,5] \Rightarrow 1\leq x \leq 5 \Rightarrow 5\geq 6-x \geq 1 \Rightarrow 1\leq \frac{5}{6-x}\leq 5$$ or $$x\in[1,5] \Rightarrow f(x)\in[1,5]$$ Now, $a_1\in[1,5] \Rightarrow a_2=f(a_1)\in[1,5]$ and, again by induction, $a_n\in[1,5]$. So, the sequence is bounded in monotone.

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  • $\begingroup$ By descending I assume you mean decreasing? $\endgroup$
    – squenshl
    Sep 28, 2019 at 9:43
  • $\begingroup$ @squenshl yes, decreasing. $\endgroup$
    – rtybase
    Sep 28, 2019 at 9:58
  • $\begingroup$ Thank you very much. Can you just go through the induction to show the sequence is decreasing? $\endgroup$
    – squenshl
    Sep 29, 2019 at 3:04
  • $\begingroup$ @squenshl it's in the 2nd and the 3rd sentences. The 2nd is "$P(1)$" step and the 3rd "$P(n) \Rightarrow P(n+1)$". $\endgroup$
    – rtybase
    Sep 29, 2019 at 9:33
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The recuurence relation is $$u_{n+1}(u_n-6)=-5~~~(1)$$ Let $$u_{n}-6=\frac{v_{n-1}}{v_{n-2}}~ in ~(1).$$ We get $$v_n+6 u_{n-1}+5 v_{n-2}~~~(2)$$ Let $v_n=x^n$ in (2), we get $$x^2+6x+5=0 \Rightarrow x_1=-5,x_2=-1$$; then $$v_n=C_{1} (-5)^n+ C_2(-1)^n~~~(3)$$ $$u_n=\frac{C_1 (-5)^{n-1}+ C_2(-1)^{n-1}}{C_1 (-5)^{n-2}+ C_2(-1)^{n-2}}+6.$$ The single unknown $D=C_1/ C_2$ can be determined by the initial values of $u_n,~$ namely $u_1=4$. Here $D$ coes out to be $5/3$, then $$U_n=6+\frac{5(-5)^{n-1}+3(-1)^{n-1}}{5(-5)^{n-2} +3 (-1)^{n-2}}.$$ ehere it follows that $\lim_{n \rightarrow \infty} u_n=1.$

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    $\begingroup$ It can be written as $$\frac{a_{n+1}-1}{a_{n+1}-5}=\frac{1}{5}\cdot\frac{a_n-1}{a_n-5}$$ Then it has form as $b_{n+1}=b_n\cdot\frac{1}{5}$ which is geometric sequence and can be solved easily $\endgroup$ Jul 4, 2020 at 20:42
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    $\begingroup$ @Very good handling of this particular case. $\endgroup$
    – Z Ahmed
    Jul 5, 2020 at 3:13

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