A rational map approaches two · source reference

Printed question

A sequence $u_1,u_2,u_3,\ldots$ is such that $u_1=1$ and

$$u_{n+1}=\frac{5u_n+4}{2u_n+3}\quad(n\ge1).$$

As $n\to\infty$, $u_n\to l$.

(i) Find $l$ exactly. [2]

(ii) Considering $y=\frac{5x+4}{2x+3}-x$, show $u_{n+1}<u_n$ if $u_n>l$. [3]

(iii) State the behaviour when $u_1=0$. [1]

Printed working

(i) As $n\to\infty$, $u_n\to l,u_{n+1}\to l$:

$$l=\frac{5l+4}{2l+3}\Rightarrow2l^2-2l-4=0\Rightarrow l^2-l-2=0\Rightarrow(l-2)(l+1)=0.$$

$l=2$ or $-1$ (rejected as $l>0$).

(ii) If $u_n>l=2$, then $y<0$ for $y=\frac{5x+4}{2x+3}-x$.

$$\frac{5u_n+4}{2u_n+3}-u_n<0\Rightarrow u_{n+1}-u_n<0\Rightarrow u_{n+1}<u_n.$$

Alternative solution:

$$u_{n+1}-u_n=\frac{5u_n+4}{2u_n+3}-u_n=\frac{-2u_n^2+2u_n+4}{2u_n+3}=\frac{-2[(u_n-1/2)^2-9/4]}{2u_n+3}<0\quad(u_n>l).$$

Hence $u_{n+1}<u_n$.

(iii) From GC, $u_n\to2$ as $n\to\infty$.

Source diagrams redrawn

Printed answer key

(i) 2; (iii) $u_n\to2$ as $n\to\infty$

Mathematical review and qualifications

(i) $2l^2-2l-4=0$, giving $l=2$ or $-1$. Starting at 1, positivity is preserved, so $l=2$.

(ii) $F(x)-x=-2(x-2)(x+1)/(2x+3)$. For $x>2$ this is negative, proving the required decrease. Also $F(x)-2=(x-2)/(2x+3)>0$, so it stays above 2.

(iii) Starting at 0, for $0\le x<2$ the increment is positive and $F(x)<2$. Induction gives increasing terms bounded above by 2. Monotone convergence and the fixed-point equation imply convergence to 2.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: RECURRENCE RELATIONS. [2011/TPJC/Prelim/I/2] Question PDF pages 3; solution PDF pages 7, 8. The original files remain private QA records outside this native content.