Halve the error at every step · source reference

Printed question

A sequence of real numbers $\{u_n\}$ satisfies

$$u_{n+1}=\frac12u_n+1,\quad n\in\mathbb Z^+.$$

(a)(i) Given convergence to $l$, find $l$. [2]

(ii) If $u_n<l$, show $u_n<u_{n+1}<l$. [3]

(iii) If $u_1<l$, describe the behaviour. [2]

(b) Another sequence has $v_n=u_n-2$, $n\in\mathbb Z^+$.

(i) Show $\{v_n\}$ is a GP. [3]

(ii) If $u_1=3$, show $u_n=2+(1/2)^{n-1}$. [2]

Printed working

(a)(i) When $u_n\to l$, then $u_{n+1}\to l$. Thus $l=\frac12l+1\Rightarrow l=2$.

(a)(ii) $u_n-u_{n+1}=u_n-(\frac12u_n+1)=\frac12u_n-1<0$ if $u_n<2$, so $u_n<u_{n+1}$.

$u_{n+1}-2=\frac12u_n+1-2=\frac12u_n-1<0$ if $u_n<2$, so $u_{n+1}<2$.

Therefore $u_n<u_{n+1}<2$ if $u_n<2$.

(a)(iii) The sequence increases and converges to 2.

(b)(i) The printed ratio proof is

$$\frac{v_{n+1}}{v_n}=\frac{u_{n+1}-2}{u_n-2}=\frac{\frac12u_n+1-2}{u_n-2}=\frac12\quad\text{(constant)},\quad n\in\mathbb Z^+.$$

(b)(ii) If $u_1=3$, $v_1=u_1-2=1$.

$$v_n=1(1/2)^{n-1}\Rightarrow u_n-2=1(1/2)^{n-1}\Rightarrow u_n=2+(1/2)^{n-1}.$$

Printed answer key

(a)(i) 2; (iii) increases and converges to 2

Mathematical review and qualifications

(a)(i) $l=l/2+1$ gives $l=2$. (ii) $u_{n+1}-u_n=(2-u_n)/2>0$ and $u_{n+1}-2=(u_n-2)/2<0$. (iii) The sequence is increasing, bounded above by 2, and converges to 2.

(b)(i) $v_{n+1}=u_{n+1}-2=(u_n-2)/2=v_n/2$, a GP relation including the zero sequence. (ii) If $u_1=3$, $v_1=1$, so $v_n=(1/2)^{n-1}$ and $u_n=2+(1/2)^{n-1}$.

Use $v_{n+1}=v_n/2$ rather than dividing by $v_n$, which could be zero.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: RECURRENCE RELATIONS. [2011/TJC/Promo/14] Question PDF pages 3; solution PDF pages 8, 9. The original files remain private QA records outside this native content.