| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Integral bounds for series |
| Difficulty | Challenging +1.8 This question requires understanding integral bounds for series approximation, integration by parts of (ln x)/x², and careful manipulation of inequalities. While the technique is standard in Further Maths, it demands multi-step reasoning across calculus and series, making it moderately challenging but within reach for well-prepared students. |
| Spec | 1.08g Integration as limit of sum: Riemann sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum_{r=1}^{N} \frac{\ln r}{r^2} = \frac{\ln 2}{4} + \sum_{r=3}^{N} \frac{\ln r}{r^2}\) | M1 A1 | Compares with sum of areas of rectangles. M1 for writing out sum, A1 for considering \(\sum_{r=3}^{N} \frac{\ln r}{r^2}\) |
| \(< \frac{\ln 2}{4} + \int_2^N \frac{\ln x}{x^2}\,dx\) | M1 | Compares with integral |
| \(\int_2^N \frac{\ln x}{x^2}\,dx = \left[-\frac{\ln x + 1}{x}\right]_2^N\) | M1 A1 | Finds integral |
| \(\sum_{r=1}^{N} \frac{\ln r}{r^2} < \frac{\ln 2}{4} + \left(-\frac{\ln N+1}{N} + \frac{\ln 2+1}{2}\right) = \frac{2+3\ln 2}{4} - \frac{1+\ln N}{N}\) | M1 A1 | Inserts limits, AG. A1 requires second M1 |
| 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum_{r=1}^{N} \frac{\ln r}{r^2} = \sum_{r=2}^{N-1} \frac{\ln r}{r^2} + \frac{\ln N}{N^2} > \int_2^N \frac{\ln x}{x^2}\,dx + \frac{\ln N}{N^2}\) | M1 A1 | Compares with integral. Lower limit of 1 scores M0. Accept \(\sum_{r=1}^{N} \frac{\ln r}{r^2} > \int_2^{N+1} \frac{\ln x}{x^2}\) |
| \(= \frac{\ln 2+1}{2} - \frac{\ln N+1}{N} + \frac{\ln N}{N^2}\) | A1 | Accept \(\frac{\ln 2+1}{2} - \frac{\ln(N+1)+1}{N+1}\) |
| 3 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum_{r=1}^{N} \frac{\ln r}{r^2} = \frac{\ln 2}{4} + \sum_{r=3}^{N} \frac{\ln r}{r^2}$ | M1 A1 | Compares with sum of areas of rectangles. M1 for writing out sum, A1 for considering $\sum_{r=3}^{N} \frac{\ln r}{r^2}$ |
| $< \frac{\ln 2}{4} + \int_2^N \frac{\ln x}{x^2}\,dx$ | M1 | Compares with integral |
| $\int_2^N \frac{\ln x}{x^2}\,dx = \left[-\frac{\ln x + 1}{x}\right]_2^N$ | M1 A1 | Finds integral |
| $\sum_{r=1}^{N} \frac{\ln r}{r^2} < \frac{\ln 2}{4} + \left(-\frac{\ln N+1}{N} + \frac{\ln 2+1}{2}\right) = \frac{2+3\ln 2}{4} - \frac{1+\ln N}{N}$ | M1 A1 | Inserts limits, AG. A1 requires second M1 |
| | **7** | |
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum_{r=1}^{N} \frac{\ln r}{r^2} = \sum_{r=2}^{N-1} \frac{\ln r}{r^2} + \frac{\ln N}{N^2} > \int_2^N \frac{\ln x}{x^2}\,dx + \frac{\ln N}{N^2}$ | M1 A1 | Compares with integral. Lower limit of 1 scores M0. Accept $\sum_{r=1}^{N} \frac{\ln r}{r^2} > \int_2^{N+1} \frac{\ln x}{x^2}$ |
| $= \frac{\ln 2+1}{2} - \frac{\ln N+1}{N} + \frac{\ln N}{N^2}$ | A1 | Accept $\frac{\ln 2+1}{2} - \frac{\ln(N+1)+1}{N+1}$ |
| | **3** | |
---\begin{enumerate}[label=(\alph*)]
\item By considering the sum of the areas of these rectangles, show that
$$\sum _ { r = 1 } ^ { N } \frac { \ln r } { r ^ { 2 } } < \frac { 2 + 3 \ln 2 } { 4 } - \frac { 1 + \ln N } { N }$$
\item Use a similar method to find, in terms of $N$, a lower bound for $\sum _ { \mathrm { r } = 1 } ^ { \mathrm { N } } \frac { \ln \mathrm { r } } { \mathrm { r } ^ { 2 } }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q4 [10]}}