RI 2025 H2 Math Prelim Paper 2 — Recurrence Relations (worked solution)

\textbf{(b)} $k = 2$ or $k = 7$. \textbf{(c)} $-1 < a < 1$, $a \neq 0$. | $L = \dfrac{b}{a}$. \textbf{(d)} $-7 < b < 7$, $b\neq 0$. | $\displaystyle\sum_{n=1}^{\infty}w_n$ is \textbf{not convergent}.

Recurrence Relations: Rational recurrence: behavior for two initial values — RI 2025 H2 Math Prelim Paper 2

Raffles Institution2025 PrelimPaper 2●●● Challenging10 marks

What this question tests

Rational recurrence: behavior for two initial values.

Question

The terms of the sequence $U$ are given by \[ u_1 = k \quad \text{and} \quad u_{n+1} = \frac{8u_n - 14}{u_n - 1},\quad n \geq 1. \]

For the following values of $k$, describe the behaviour of the sequence $U$.

$k = 3$

$k = 10$

Find the possible value(s) of $k$ if $U$ is a constant sequence.

The $n$th term of the sequence $V$ is given by $v_n = \dfrac{a}{b} \cdot \dfrac{b}{a(1-a^n)+1} \cdot \dfrac{a}{n+1}$, where $a$ and $b$ are non-zero real constants and $a \neq \pm 1$.

For some values of $a$, $v_n \to L$ as $n \to \infty$. Find, with justification, the range of values of $a$ for $L$ to exist, and state the value of $L$ in terms of $a$ and $b$.

The $n$th term of the sequence $W$ is given by \[ w_n = \begin{cases} v_n & \text{when } n \text{ is even,}\\ u_n & \text{when } n \text{ is odd.}\end{cases} \] It is given that the sequence $W$ converges when the sequences $U$ and $V$ converge to the same limit. The sequence $W$ diverges otherwise.

For $k = 10$, by using part (a)(ii) and part (c), find the range of values of $b$ for sequence $W$ to converge. Hence explain whether $\displaystyle\sum_{n=1}^{\infty} w_n$ is a convergent series.

Show full worked solution▾

From GC:

(i) For $k = 3$, the terms are increasing and converging to $7$.

(ii) For $k = 10$, the terms are decreasing and converging to $7$.

For a constant sequence, $u_{n+1} = u_n = k$: \[\begin{aligned} k &= \frac{8k-14}{k-1}\\[4pt] k(k-1) &= 8k-14\\[4pt] k^2 - k &= 8k - 14\\[4pt] k^2 - 9k + 14 &= 0\\[4pt] (k-2)(k-7) &= 0 \end{aligned}\]

\[ v_n = \frac{a^n}{b} + \frac{b}{a(1-a^n)} - \frac{a}{n+1} \]

As $n \to \infty$, $\dfrac{1}{n+1} \to 0 \implies \dfrac{a}{n+1} \to 0$ for any constant $a \neq 0,\, \pm 1$.

If $|a| > 1$, as $n \to \infty$, $\left|\dfrac{a^n}{b}\right|$ increases without bound, which implies $V$ is not convergent.

If $-1 < a < 1$ and $a \neq 0$, as $n \to \infty$, $a^n \to 0$.

Hence for $V$ to be convergent, $-1 < a < 1$ and $a \neq 0$.

The required range of values of $a$ is $(-1,0) \cup (0,1)$ and the limiting value $L$ is $\dfrac{b}{a}$.

From (a)(ii), $u_n \to 7$, and from (c), $L = \dfrac{b}{a}$ for $-1 < a < 1$, $a \neq 0$.

Hence, since W converges, \[\begin{aligned} \frac{b}{a} &= 7 \text{ and } -1 < a < 1 \text{ and } a \neq 0 \\ \implies b &= 7a \text{ and } -1 < a < 1 \text{ and } a \neq 0 \end{aligned}\]

Since $-1 < a < 1$ and $a \neq 0$, $-7 < b < 0$ or $0 < b < 7$.

$\implies$ Range of values of $b$ for W to converge is $(-7, 0) \cup (0, 7)$.

Since the limiting value of the sequence W is a non-zero value (7), the sum of an infinite number of non-zero values is arbitrarily large and cannot converge to a particular value. Hence the series $\displaystyle\sum_{r=1}^{\infty} w_r$ is not convergent.

Answer: (b) $k = 2$ or $k = 7$. (c) $-1 < a < 1$, $a \neq 0$. $L = \dfrac{b}{a}$. (d) $-7 < b < 7$, $b\neq 0$. $\displaystyle\sum_{n=1}^{\infty}w_n$ is not convergent.

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