Year: 2024
Paper: 3
Question Number: 1
Course: UFM Additional Further Pure
Section: Sequences and Series
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
Throughout this question, $N$ is an integer with $N \geqslant 1$ and $S_N = \displaystyle\sum_{r=1}^{N} \frac{1}{r^2}$.
You may assume that $\displaystyle\lim_{N\to\infty} S_N$ exists and is equal to $\frac{1}{6}\pi^2$.
\begin{questionparts}
\item Show that
\[\frac{1}{r+1} - \frac{1}{r} + \frac{1}{r^2} = \frac{1}{r^2(r+1)}.\]
Hence show that
\[\sum_{r=1}^{N} \frac{1}{r^2(r+1)} = \sum_{r=1}^{N} \frac{1}{r^2} - 1 + \frac{1}{N+1}.\]
Show further that $\displaystyle\sum_{r=1}^{\infty} \frac{1}{r^2(r+1)} = \frac{1}{6}\pi^2 - 1$.
\item Find $\displaystyle\sum_{r=1}^{N} \frac{1}{r^2(r+1)(r+2)}$ in terms of $S_N$, and hence evaluate $\displaystyle\sum_{r=1}^{\infty} \frac{1}{r^2(r+1)(r+2)}$.
\item Show that
\[\sum_{r=1}^{\infty} \frac{1}{r^2(r+1)^2} = \sum_{r=1}^{\infty} \frac{2}{r^2(r+1)} - 1.\]
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& \frac1{r+1} - \frac1r + \frac1{r^2} &= \frac{-1}{r(r+1)} + \frac{1}{r^2} \\
&&&= \frac{r+1-r}{r^2(r+1)} \\
&&&= \frac{1}{r^2(r+1)}
\end{align*}
Therefore
\begin{align*}
&& \sum_{r=1}^N \frac1{r^2(r+1)} &= \sum_{r=1}^N \left (\frac1{r+1} - \frac1r + \frac1{r^2} \right) \\
&&&= \sum_{r=1}^N \left (\frac1{r+1} - \frac1r\right) + \sum_{r=1}^N \frac1{r^2} \\
&&&=\frac{1}{N+1} - 1 + \sum_{r=1}^N \frac1{r^2} \\
\end{align*}
therefore
\begin{align*}
&& \sum_{r=1}^{\infty} \frac{1}{r^2(r+1)} &= \lim_{N \to \infty } \sum_{r=1}^{N} \frac{1}{r^2(r+1)} \\
&&&= \lim_{N \to \infty } \left (\frac{1}{N+1} - 1 + \sum_{r=1}^N \frac1{r^2} \right) \\
&&&= -1 +\lim_{N \to \infty } \sum_{r=1}^N \frac1{r^2} \\
&&&= -1 + \sum_{r=1}^\infty \frac1{r^2} \\
&&&= \frac{\pi^2}{6}-1
\end{align*}
\item Note that \begin{align*}
&& \frac{1}{r^2(r+1)(r+2)} &= \frac{Ar+B}{r^2} + \frac{C}{r+1} + \frac{D}{r+2} \\
&&&= \frac{1}{2r^2} + \frac{1}{r+1} - \frac{1}{4(r+2)} - \frac{3}{4r}
\end{align*}
So
\begin{align*}
&& \sum_{r=1}^N \frac{1}{r^2(r+1)(r+2)} &= \sum_{r=1}^N \left ( \frac{1}{2r^2} + \frac{1}{r+1} - \frac{1}{4(r+2)} - \frac{3}{4r} \right ) \\
&&&= \frac12 \sum_{r=1}^N \frac{1}{r^2} + \frac{1}{2} - \frac14 \cdot \frac1{3} - \frac34 \frac11 + \\
&&& \quad \quad \quad + \frac13 - \frac14\frac14 - \frac34\frac12 + \\
&&& \quad \quad \quad + \frac14 - \frac14\frac15 - \frac34\frac13 + \\
&&&\quad \quad \quad+ \cdots + \\
&&&\quad \quad \quad + \frac1{N+1} - \frac14\frac1{N+2} - \frac34\frac1N \\
&&&= \frac12 \sum_{r=1}^N \frac{1}{r^2} +\frac14\frac12 - \frac34\frac11-\frac14\frac1{N+2}+\frac34 \frac1{N+1} \\
\\
\Rightarrow && \sum_{r=1}^{\infty} \frac{1}{r^2(r+1)(r+2)} &= \lim_{N \to \infty} \left [ \frac12 \sum_{r=1}^N \frac{1}{r^2} +\frac14\frac12 - \frac34\frac11-\frac14\frac1{N+2}+\frac34 \frac1{N+1}\right] \\
&&&= \frac{\pi^2}{12} -\frac58
\end{align*}
\item Notice that $\frac{1}{r^2(r+1)^2} - \frac{2}{r^2(r+1)} = \frac{1-2(r+1)}{r^2(r+1)^2} = \frac{-1-2r}{r^2(r+1)^2} = \frac{1}{(r+1)^2} - \frac{1}{r^2}$ and so
\begin{align*}
&& \sum_{r=1}^N \frac{1}{r^2(r+1)^2} &= \sum_{r=1}^N \left ( \frac{2}{r^2(r+1)} +\frac{1}{(r+1)^2} - \frac{1}{r^2} \right) \\
&&&= \sum_{r=1}^N \frac{2}{r^2(r+1)} +\frac{1}{(N+1)^2} - 1 \\
\end{align*}
and the result follows as $N \to \infty$
\end{questionparts}
[There is a beautiful paper by KConrad about this question: https://kconrad.math.uconn.edu/blurbs/analysis/series_acceleration.pdf]
This was comfortably both the most popular question and the most successful, with a mean score of about 15/20. There were numerous correct methods employed to approach the partial fractions. Every part had many excellent clear responses. Generally, if candidates could do the partial fractions algorithm correctly and wrote more than the bare minimum for the limiting and telescoping operations they got almost full marks. In part (i), most could do the calculations correctly, though explanations less so. In parts (ii) and (iii), many candidates did not attempt the correct decomposition. Explanations of cancelling terms in the telescoping series and taking limits were frequently not clear. Particular weaknesses were treating harmonic series as if they converged, and substituting ∞ into expressions as if it were a number. There were many clever ways of doing the last part without a full partial fraction decomposition, but probably the cleanest was as follows.