| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Integral bounds for series |
| Difficulty | Standard +0.8 This is a well-structured FP2 question on integral inequalities and the harmonic series. Part (i) requires understanding that rectangles under the curve give an underestimate of the integral (5 marks suggests detailed explanation needed). Parts (ii-iii) follow logically with overestimate rectangles and algebraic manipulation. Part (iv) tests understanding of divergence. While requiring careful geometric reasoning and proof writing typical of Further Maths, the concepts are standard for FP2 and the question guides students through each step systematically. |
| Spec | 1.04j Sum to infinity: convergent geometric series |r|<11.06d Natural logarithm: ln(x) function and properties1.08g Integration as limit of sum: Riemann sums |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Area \(= \int \frac{1}{l(x+1)} dx\). Use limits to \(\ln(n+1)\). Compare area under curve to areas of rectangles. Sum of areas of rectangles \(= 1 \cdot (\frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n+1})\). Clear detail to A.G. | B1, B1, B1, M1, A1 | Include or imply correct limits; Justify inequality; Sum seen or implied as 1 x y values; Explanation required e.g. area of last rectangle at \(x=n\), area under curve to \(x=n\) |
| (ii) Show or explain areas of rectangles above curve. Areas of rectangles (as above) > area under curve | M1, A1 | First and last heights seen or implied; A.G. |
| (iii) Add 1 to both sides in (i) to make \(\sum(\frac{1}{r})\). Add \(\frac{1}{(n+1)}\) to both sides in (ii) to make \(\sum(\frac{1}{r})\) | B1, B1 | Must be clear addition; Must be clear addition; A.G. |
| (iv) State divergent. Explain e.g. \(\ln(n+1) \to \infty\) as \(n \to \infty\) | B1, B1 | Allow not convergent |
(i) Area $= \int \frac{1}{l(x+1)} dx$. Use limits to $\ln(n+1)$. Compare area under curve to areas of rectangles. Sum of areas of rectangles $= 1 \cdot (\frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n+1})$. Clear detail to A.G. | B1, B1, B1, M1, A1 | Include or imply correct limits; Justify inequality; Sum seen or implied as 1 x y values; Explanation required e.g. area of last rectangle at $x=n$, area under curve to $x=n$
(ii) Show or explain areas of rectangles above curve. Areas of rectangles (as above) > area under curve | M1, A1 | First and last heights seen or implied; A.G.
(iii) Add 1 to both sides in (i) to make $\sum(\frac{1}{r})$. Add $\frac{1}{(n+1)}$ to both sides in (ii) to make $\sum(\frac{1}{r})$ | B1, B1 | Must be clear addition; Must be clear addition; A.G.
(iv) State divergent. Explain e.g. $\ln(n+1) \to \infty$ as $n \to \infty$ | B1, B1 | Allow not convergent\includegraphics{figure_8}
The diagram shows the curve with equation $y = \frac{1}{x+1}$. A set of $n$ rectangles of unit width is drawn, starting at $x = 0$ and ending at $x = n$, where $n$ is an integer.
\begin{enumerate}[label=(\roman*)]
\item By considering the areas of these rectangles, explain why
$$\frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n+1} < \ln(n+1).$$ [5]
\item By considering the areas of another set of rectangles, show that
$$1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n} > \ln(n+1).$$ [2]
\item Hence show that
$$\ln(n+1) + \frac{1}{n+1} < \sum_{r=1}^{n+1} \frac{1}{r} < \ln(n+1) + 1.$$ [2]
\item State, with a reason, whether $\sum_{r=1}^{\infty} \frac{1}{r}$ is convergent. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR FP2 2009 Q8 [11]}}