| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2008 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Method of differences with logarithmic terms |
| Difficulty | Standard +0.8 This is a telescoping series question requiring recognition that logarithm properties allow consecutive terms to cancel. While the technique is standard for Further Maths, students must handle logarithmic manipulation, find the general sum formula, then carefully analyze limits for different cases of x. The multi-part structure and need to consider convergence conditions elevates it above routine exercises. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules4.06b Method of differences: telescoping series |
| Answer | Marks |
|---|---|
| \(u_n = \ln(1 + x^{n+1}) - \ln(1 + x)\) or for \(\ln\{\text{Product of fractions}\}\) | B1 |
| \(\sum_{n=1}^N u_n = S_N = \ln[(1 + x^{N+1})(1 + x)]\) (AEF) Cancels \(\to\) result | M1A1 |
| (i) \(S_e = -\ln(1 + x)\) OR \(\ln\left(\frac{1}{1+x}\right)\) | A1 |
| (ii) \(S_\infty = 0\) | B1 |
$u_n = \ln(1 + x^{n+1}) - \ln(1 + x)$ or for $\ln\{\text{Product of fractions}\}$ | B1 |
$\sum_{n=1}^N u_n = S_N = \ln[(1 + x^{N+1})(1 + x)]$ (AEF) Cancels $\to$ result | M1A1 |
**(i)** $S_e = -\ln(1 + x)$ **OR** $\ln\left(\frac{1}{1+x}\right)$ | A1 |
**(ii)** $S_\infty = 0$ | B1 |2 Given that
$$u _ { n } = \ln \left( \frac { 1 + x ^ { n + 1 } } { 1 + x ^ { n } } \right)$$
where $x > - 1$, find $\sum _ { n = 1 } ^ { N } u _ { n }$ in terms of $N$ and $x$.
Find the sum to infinity of the series
$$u _ { 1 } + u _ { 2 } + u _ { 3 } + \ldots$$
when\\
(i) $- 1 < x < 1$,\\
(ii) $x = 1$.
\hfill \mbox{\textit{CAIE FP1 2008 Q2 [5]}}