Year: 2013
Paper: 2
Question Number: 6
Course: UFM Pure
Section: Sequences and series, recurrence and convergence
All questions were attempted by a significant number of candidates, with questions 1 to 3 and 7 the most popular. The Pure questions were more popular than both the Mechanics and the Probability and Statistics questions, with only question 8 receiving a particularly low number of attempts within the Pure questions and only question 11 receiving a particularly high number of attempts.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1485.5
Banger Comparisons: 1
In this question, the following theorem may be used.
\textit{Let $u_1$, $u_2$, $\ldots$ be a sequence of (real) numbers. If the sequence is bounded above (that is, $u_n\le b$ for all $n$, where $b$ is some fixed number) and increasing (that is, }$u_n \ge u_{n-1}$\textit{ for all $n$), then the sequence tends to a limit (that is, converges).}
The sequence $u_1$, $u_2$, $\ldots$ is defined by $u_1=1$ and
\[
u_{n+1} = 1+\frac 1{u_n} \ \ \ \ \ \ \ \ \ \ (n\ge1)\,.
\tag{$*$}
\]
\begin{questionparts}
\item Show that, for $n\ge3$,
\[
u_{n+2}-u_n = \frac{u_{n} - u_{n-2}}{(1+u_n)(1+u_{n-2})}
.
\]
\item Prove, by induction or otherwise, that $1\le u_n \le 2$ for all $n$.
\item
Show that the sequence $u_1$, $u_3$, $u_5$, $\ldots$ tends to a limit, and that the sequence $u_2$, $u_4$, $u_6$, $\ldots$
tends to a limit. Find these limits and deduce that the sequence $u_1$, $u_2$, $u_3$, $\ldots\,$ tends to a limit.
Would this conclusion change if the sequence were defined by $(*)$ and $u_1=3$?
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& u_{n+2} - u_n &= 1 + \frac{1}{u_{n+1}} - \left (1 + \frac{1}{u_{n-1}} \right) \\
&&&= \frac{1}{1 + \frac1{u_n}} - \frac{1}{1 + \frac{1}{u_{n-2}}} \\
&&&= \frac{u_n}{u_n+1} - \frac{u_{n-2}}{1+u_{n-2}} \\
&&&= \frac{u_n(1+u_{n-2}) - u_{n-2}(1+u_n)}{(1+u_n)(1+u_{n-2})} \\
&&&= \frac{u_n - u_{n-2}}{(1+u_n)(1+u_{n-2})}
\end{align*}
\item Claim: $u_n \in [1,2]$
Proof: (By induction).
Note that $u_1 = 1, u_2 = 2$ so our claim is true for the first few terms.
Note that if $u_n \in [1,2]$, $\frac{1}{u_n} \in [\tfrac12, 1]$ and $1+\frac{1}{u_{n}} \in [\tfrac32,2] \subset [1,2]$. Therefore $u_{n+1} \in [1,2]$.
Therefore since $u_1 \in [1,2]$ and $u_n \in [1,2] \Rightarrow u_{n+1} \in [1,2]$ $u_n \in [1,2]$ for all $n \ge 1$
\item First notice that $u_3 = \frac32 > u_1$ and therefore by the recursion we found in the first part, $u_{2n+1}-u_{2n-1} > 0$ so $u_{2k+1}$ is increasing and bounded, and so by our theorem converges to a limit. Suppose this limit is $L$, then we must have $L = 1 + \frac1{L} \Rightarrow L^2 - L - 1 = 0 \Rightarrow L = \frac{1+\sqrt5}{2}$ since it must be in $[1,2]$.
Similarly, not that $u_4 = \frac{5}{3} < u_2$ and so $u_{2k+2} - u_{2k} < 0$ and $-u_{2k}$ is increasing and bounded above. Therefore it tends to a limit (and so does $u_{2k}$). By the same reasoning as before, it's the same limit, $\frac{1+\sqrt5}{2}$ and therefore the sequence converges.
If $u_1 = 3, u_2 = \frac43 \in [1,2]$ so we have our sequence being bounded and all the same logic will follow through.
\end{questionparts}
The algebra required for the first part of the question proved to be quite challenging for a number of candidates, but most were able to reach the required answer. The proof by induction in the second part of the question was generally well done, although a number of candidates did not write up the process clearly. In the final part of the question it was clear that many candidates had identified the relationship between the sequences and Fibonacci numbers and some candidates therefore stated that the limit would be the golden ratio, but without any supporting calculations. In the final part there were few responses which clearly explained that the new sequence would still satisfy the conditions required if it were started at a later term.