| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2013 |
| Session | November |
| Marks | 10 |
| Topic | Sequences and series, recurrence and convergence |
| Type | Integral bounds for series |
| Difficulty | Challenging +1.3 This is a structured Further Maths question on series that guides students through method of differences (a standard technique), taking limits, and establishing bounds. Part (i)(a) is routine application of partial fractions and telescoping. Part (i)(b) requires recognizing the limit and making a comparison, which is moderately challenging. Part (ii) applies these results to bound a famous series, requiring careful bookkeeping but following naturally from part (i). The scaffolding and standard techniques place this above average difficulty but well within reach for prepared Further Maths students. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series |
\begin{enumerate}[label=(\roman*)]
\item \begin{enumerate}[label=(\alph*)]
\item Use the method of differences to prove that
$$\sum_{n=k}^N \frac{1}{n(n+1)} = \frac{1}{k} - \frac{1}{N+1}.$$ [4]
\item Deduce the value of $\sum_{n=k}^{\infty} \frac{1}{n(n+1)}$ and show that $\sum_{n=k}^{\infty} \frac{1}{(n+1)^2} < \frac{1}{k}$. [3]
\end{enumerate}
\item Let $S = \sum_{n=1}^{\infty} \frac{1}{n^2}$. Show that $\frac{205}{144} < S < \frac{241}{144}$. [3]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2013 Q12 [10]}}