Recurrence Relations: Factorial recurrence: general term & series convergence — NJC 2025 H2 Math Prelim Paper 1
What this question tests
Question
A sequence of numbers \(u_1,u_2,u_3,\ldots\) has a sum \(S_n\), where \(S_n=\displaystyle\sum_{r=1}^n u_r\). It is given that \[ S_n = A - \frac{2}{(n+1)!}, \] where \(A\) is a non-zero constant.
- Find the value of \(A\) if \(u_1=1\).
- Show that \(u_n=\dfrac{2}{(n+1)\bigl[(n-1)!\bigr]}\) for \(n\geq 1\).
- Find a recurrence relation in the form \(u_{n+1}=\bigl[\mathrm{f}(n)\bigr]u_n\).
- Explain why \(S_n\) converges as \(n\to\infty\).
- Hence, find the least value of \(m\) such that the sum of the infinite series \[ u_m+u_{m+1}+u_{m+2}+\cdots \] does not exceed \(10^{-10}\).
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(i)
\(u_1 = S_1 = A - \dfrac{2}{2!} = A-1\).
Given \(u_1=1\): \(A-1=1\).
\(A=2\)(ii)
For \(n\geq 2\): \[\begin{aligned} u_n &= S_n - S_{n-1} = \!\left[2-\frac{2}{(n+1)!}\right]-\!\left[2-\frac{2}{n!}\right] \\ &= \frac{2}{n!}-\frac{2}{(n+1)!} \\ &= \frac{2(n+1)}{(n+1)!}-\frac{2}{(n+1)!} \\ &= \frac{2n}{(n+1)!} \\ &= \frac{2n}{(n+1)(n)(n-1)!} \\ &= \frac{2}{(n+1)[(n-1)!]} \end{aligned}\]
Check for \(n=1\): \(\dfrac{2}{(2)(0!)} = \dfrac{2}{2(1)} = 1 = u_1\).
Hence \(u_n = \dfrac{2}{(n+1)[(n-1)!]}\) for \(n\geq 1\). \( \)
(iii) \[\begin{aligned} u_{n+1} &= \frac{2}{(n+2)(n!)} \\ \frac{u_{n+1}}{u_n} &= \frac{2}{(n+2)(n!)} \div \frac{2}{(n+1)[(n-1)!]} = \frac{(n+1)[(n-1)!]}{(n+2)(n!)} = \frac{n+1}{n(n+2)} \end{aligned}\]
\(u_{n+1} = \dfrac{n+1}{n(n+2)}\,u_n\)(iv)
As \(n\to\infty\), \((n+1)!\to\infty\), so \(\dfrac{2}{(n+1)!}\to 0\).
Therefore \(S_n = 2-\dfrac{2}{(n+1)!} \to 2\).
\(S_n\) converges to 2.
(v) \[\begin{aligned} u_m+u_{m+1}+\cdots &= \sum_{r=1}^\infty u_r - \sum_{r=1}^{m-1} u_r = 2 - S_{m-1} \\ &= 2 - \!\left[2-\frac{2}{m!}\right] = \frac{2}{m!} \leq 10^{-10} \end{aligned}\]
By GC:
| \(m\) | \(\dfrac{2}{m!}\) |
|---|---|
| 13 | \(3.21\times10^{-10}\) |
| 14 | \(2.29\times10^{-11}\) |