Year: 2024
Paper: 3
Question Number: 7
Course: UFM Additional Further Pure
Section: Sequences and Series
No solution available for this problem.
The total entry was an increase on that of 2023 by more than 10%. One question was attempted by more than 98% of candidates, another two by about 80%, and another five by between 50% and 70%. The remaining four questions were attempted by between 5% and 30% of candidates, these being from Section B: Mechanics, and Section C: Probability and Statistics, though the Statistics questions were in general attempted more often and more successfully. All questions were perfectly solved by some candidates. About 84% of candidates attempted no more than 7 questions.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
In this question, you need not consider issues of convergence.
For positive integer $n$ let
\[\mathrm{f}(n) = \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + \ldots\]
and
\[\mathrm{g}(n) = \frac{1}{n+1} - \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} - \ldots\,.\]
\begin{questionparts}
\item Show, by considering a geometric series, that $0 < \mathrm{f}(n) < \dfrac{1}{n}$.
\item Show, by comparing consecutive terms, that $0 < \mathrm{g}(n) < \dfrac{1}{n+1}$.
\item Show, for positive integer $n$, that $(2n)!\,\mathrm{e} - \mathrm{f}(2n)$ and $\dfrac{(2n)!}{\mathrm{e}} + \mathrm{g}(2n)$ are both integers.
\item Show that if $q\,\mathrm{e} = \dfrac{p}{\mathrm{e}}$ for some positive integers $p$ and $q$, then $q\,\mathrm{f}(2n) + p\,\mathrm{g}(2n)$ is an integer for all positive integers $n$.
\item Hence show that the number $\mathrm{e}^2$ is irrational.
\end{questionparts}
The third most popular question, this was a little less successfully attempted than question 2 with a mean score of just over 9/20. Parts (i) and (ii) were not generally well done, as it was easy to guess the geometric series and then make unsubstantiated, or at least unjustified, claims which could not be given full marks. In part (ii), there was frequently lack of clarity regarding pairing of terms and arguments lacking in necessary detail to support the claims. Part (iii) was done better, though the second result commonly saw 1/e expanded as a reciprocal rather than as e-1, and then, as a consequence, getting lost. Part (iv), too, was fairly well done. There was a good understanding of contradiction arguments for part (v), though there was difficulty in choosing a suitable n in quite a few cases.