Year: 2021
Paper: 3
Question Number: 8
Course: UFM Pure
Section: Sequences and series, recurrence and convergence
The total entry was a marginal increase from that of 2019, that of 2020 having been artificially reduced. Comfortably more than 90% attempted one of the questions, four others were very popular, and a sixth was attempted by 70%. Every question was attempted by at least 10% of the candidature. 85% of candidates attempted no more than 7 questions, though very nearly all the candidates made genuine attempts on at most six questions (the extra attempts being at times no more than labelling a page or writing only the first line or two). Generally, candidates should be aware that when asked to "Show that" they must provide enough working to fully substantiate their working, and that they should follow the instructions in a question, so if it says "Hence", they should be using the previous work in the question in order to complete the next part. Likewise, candidates should be careful when dividing or multiplying, that things are positive, or at other times non-zero.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
A sequence $x_1, x_2, \ldots$ of real numbers is defined by $x_{n+1} = x_n^2 - 2$ for $n \geqslant 1$ and $x_1 = a$.
\begin{questionparts}
\item Show that if $a > 2$ then $x_n \geqslant 2 + 4^{n-1}(a-2)$.
\item Show also that $x_n \to \infty$ as $n \to \infty$ if and only if $|a| > 2$.
\item When $a > 2$, a second sequence $y_1, y_2, \ldots$ is defined by
\[
y_n = \frac{Ax_1 x_2 \cdots x_n}{x_{n+1}},
\]
where $A$ is a positive constant and $n \geqslant 1$.
Prove that, for a certain value of $a$, with $a > 2$, which you should find in terms of $A$,
\[
y_n = \frac{\sqrt{x_{n+1}^2 - 4}}{x_{n+1}}
\]
for all $n \geqslant 1$.
Determine whether, for this value of $a$, the second sequence converges.
\end{questionparts}\begin{questionparts}
\item Claim $x_n \geqslant 2 + 4^{n-1}(a-2)$
Proof: (By induction)
Base case: Note that when $n = 1$, $x_1 = a = 2 + 1 \cdot(a - 2)$.
Inductive step, suppose true for some $n$, then
\begin{align*}
&& x_{n+1} &= x_n^2 - 2 \\
&&&\geq (2+4^{n-1}(a-2))^2 - 2 \\
&&&= 4 + 4^{2n-2}(a-2)^2 + 4^n(a-2) - 2 \\
&&&= 2 + 4^{n}(a-2) + 4^{2n-2}(a-2)^2 \\
&&&\geq 2 + 4^{n+1-1}(a-2)
\end{align*}
as required,
\item ($\Leftarrow$) Suppose $a > 2$ then $x_n \geq 2+4^{n-1}(a-2) \to \infty$ as required. Suppose $a < -2$ then $x_2 > 4 -2 = 2$ so the sequence starting from $x_2$ clearly diverges for the same reason.
($\Rightarrow$) suppose $|x_n| \leq 2$ then $x_{n+1} = x_n^2 - 2 \leq 2$ so the sequence is bounded and cannot tend to $\infty$.
\item Suppose $y_n = \frac{Ax_1x_2 \cdots x_n}{x_{n+1}}$ and notice that $x_{n+1}^2 - 4 = (x_n^2 -2)^2 - 4 = x_n^4 - 4x_n^2 = x_n^2(x_n^2-4)$.
In particular, $\frac{\sqrt{x_{n+1}^2-4}}{x_{n+1}} = \frac{x_n\sqrt{x_n^2-4}}{x_{n+1}} = \frac{x_n x_{n-1} \cdots x_1 \sqrt{x_1^2-4}}{x_{n+1}}$ Therefore if $A = \sqrt{a^2-4}$ $y_{n+1} = \frac{\sqrt{x_{n+1}^2-4}}{x_{n+1}}$.
Notice that \begin{align*}
&& y_n &= \frac{\sqrt{x_{n+1}^2-4}}{x_{n+1}} \\
&&&= \sqrt{1 - \frac{4}{x_{n+1}^2}} \to 1
\end{align*}
\end{questionparts}
Fifth most popular (77%), this was fourth least successful with a mean mark of six and a half. There were very few perfect attempts and a sizeable number of attempts failed to get any marks. Induction in both parts (i) and (iii) was generally executed very well, however marks were frequently lost for logical imprecision. A very common cause of lost marks was a lack of care with inequalities involving potentially negative numbers. In part (i), almost no candidates noticed that squaring the inequality required noting the non-negativity of the lower bound. Many candidates also had trouble with the base case, some because they were mistakenly thinking 4⁰ = 0. In part (ii), many candidates lost marks when attempting to show that the sequence |x_n| remains bounded in the case |a| < 2, by not excluding the possibility that x₂ goes below −2 and hence diverges to positive infinity. Another common error in part (ii) was failing to make the link to the inequality in part (i). Many candidates tried to show divergence to infinity by showing that the sequence was increasing. In part (iii) most candidates worked back from the required result to find a suitable value for a. The inductive calculation was generally performed well, however plenty of candidates failed to show that their value of a worked and was greater than 2. When solving equations, it should either be checked that all the steps are reversible (in this case they were not because of a possible division by zero) or that the claimed solution does in fact work. Most attempts at the final section on convergence were informal but successful.