| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Method of differences with given identity |
| Difficulty | Standard +0.8 This is a standard Further Maths telescoping series question with three routine parts: algebraic verification, applying method of differences to find a closed form, and solving an inequality involving the tail of the series. While it requires competence with exponentials and series manipulation, all techniques are standard FP1 material with no novel insights required. |
| Spec | 4.06b Method of differences: telescoping series8.01d Sequence limits: limit of nth term as n tends to infinity, steady-states |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{ne^n} - \frac{1}{(n+1)e^{n+1}} = \frac{(n+1)e - n}{n(n+1)e^{n+1}} = \frac{n(e-1)+e}{n(n+1)e^{n+1}}\) | B1 | Verifies result (AG) |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_N = \sum_{n=1}^{N}\left(\frac{1}{ne^n} - \frac{1}{(n+1)e^{n+1}}\right) = \left(\frac{1}{e} - \frac{1}{2e^2} + \frac{1}{2e^2} - \frac{1}{3e^3} + \cdots - \frac{1}{Ne^N} - \frac{1}{(N+1)e^{N+1}}\right)\) SOI | M1 | Uses difference method to sum |
| \(\frac{1}{e} - \frac{1}{(N+1)e^{N+1}}\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S = \frac{1}{e}\) | B1 | Finds S |
| \((N+1)(S - S_N) < 10^{-3} \Rightarrow \frac{1}{e^{N+1}} < 10^{-3}\) | M1 | Attempts to find difference between sum and sum to infinity |
| \(\Rightarrow e^{N+1} > 10^3 \Rightarrow\) least such \(N\) is 6 | A1 |
**Question 2:**
**Part (i):**
$\frac{1}{ne^n} - \frac{1}{(n+1)e^{n+1}} = \frac{(n+1)e - n}{n(n+1)e^{n+1}} = \frac{n(e-1)+e}{n(n+1)e^{n+1}}$ | **B1** | Verifies result (AG)
**Part (ii):**
$S_N = \sum_{n=1}^{N}\left(\frac{1}{ne^n} - \frac{1}{(n+1)e^{n+1}}\right) = \left(\frac{1}{e} - \frac{1}{2e^2} + \frac{1}{2e^2} - \frac{1}{3e^3} + \cdots - \frac{1}{Ne^N} - \frac{1}{(N+1)e^{N+1}}\right)$ SOI | **M1** | Uses difference method to sum
$\frac{1}{e} - \frac{1}{(N+1)e^{N+1}}$ | **A1** | —
## Question 2(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S = \frac{1}{e}$ | B1 | Finds S |
| $(N+1)(S - S_N) < 10^{-3} \Rightarrow \frac{1}{e^{N+1}} < 10^{-3}$ | M1 | Attempts to find difference between sum and sum to infinity |
| $\Rightarrow e^{N+1} > 10^3 \Rightarrow$ least such $N$ is 6 | A1 | |
---2 (i) Verify that
$$\frac { n ( \mathrm { e } - 1 ) + \mathrm { e } } { n ( n + 1 ) \mathrm { e } ^ { n + 1 } } = \frac { 1 } { n \mathrm { e } ^ { n } } - \frac { 1 } { ( n + 1 ) \mathrm { e } ^ { n + 1 } }$$
Let $S _ { N } = \sum _ { n = 1 } ^ { N } \frac { n ( \mathrm { e } - 1 ) + \mathrm { e } } { n ( n + 1 ) \mathrm { e } ^ { n + 1 } }$.\\
(ii) Express $S _ { N }$ in terms of $N$ and e.\\
Let $S = \lim _ { N \rightarrow \infty } S _ { N }$.\\
(iii) Find the least value of $N$ such that $( N + 1 ) \left( S - S _ { N } \right) < 10 ^ { - 3 }$.\\
\hfill \mbox{\textit{CAIE FP1 2018 Q2 [6]}}