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2024 Paper 3 Q1
D: 1500.0 B: 1500.0

Throughout this question, \(N\) is an integer with \(N \geqslant 1\) and \(S_N = \displaystyle\sum_{r=1}^{N} \frac{1}{r^2}\). You may assume that \(\displaystyle\lim_{N\to\infty} S_N\) exists and is equal to \(\frac{1}{6}\pi^2\).

  1. Show that \[\frac{1}{r+1} - \frac{1}{r} + \frac{1}{r^2} = \frac{1}{r^2(r+1)}.\] Hence show that \[\sum_{r=1}^{N} \frac{1}{r^2(r+1)} = \sum_{r=1}^{N} \frac{1}{r^2} - 1 + \frac{1}{N+1}.\] Show further that \(\displaystyle\sum_{r=1}^{\infty} \frac{1}{r^2(r+1)} = \frac{1}{6}\pi^2 - 1\).
  2. Find \(\displaystyle\sum_{r=1}^{N} \frac{1}{r^2(r+1)(r+2)}\) in terms of \(S_N\), and hence evaluate \(\displaystyle\sum_{r=1}^{\infty} \frac{1}{r^2(r+1)(r+2)}\).
  3. Show that \[\sum_{r=1}^{\infty} \frac{1}{r^2(r+1)^2} = \sum_{r=1}^{\infty} \frac{2}{r^2(r+1)} - 1.\]


Solution:

  1. \(\,\) \begin{align*} && \frac1{r+1} - \frac1r + \frac1{r^2} &= \frac{-1}{r(r+1)} + \frac{1}{r^2} \\ &&&= \frac{r+1-r}{r^2(r+1)} \\ &&&= \frac{1}{r^2(r+1)} \end{align*} Therefore \begin{align*} && \sum_{r=1}^N \frac1{r^2(r+1)} &= \sum_{r=1}^N \left (\frac1{r+1} - \frac1r + \frac1{r^2} \right) \\ &&&= \sum_{r=1}^N \left (\frac1{r+1} - \frac1r\right) + \sum_{r=1}^N \frac1{r^2} \\ &&&=\frac{1}{N+1} - 1 + \sum_{r=1}^N \frac1{r^2} \\ \end{align*} therefore \begin{align*} && \sum_{r=1}^{\infty} \frac{1}{r^2(r+1)} &= \lim_{N \to \infty } \sum_{r=1}^{N} \frac{1}{r^2(r+1)} \\ &&&= \lim_{N \to \infty } \left (\frac{1}{N+1} - 1 + \sum_{r=1}^N \frac1{r^2} \right) \\ &&&= -1 +\lim_{N \to \infty } \sum_{r=1}^N \frac1{r^2} \\ &&&= -1 + \sum_{r=1}^\infty \frac1{r^2} \\ &&&= \frac{\pi^2}{6}-1 \end{align*}
  2. Note that \begin{align*} && \frac{1}{r^2(r+1)(r+2)} &= \frac{Ar+B}{r^2} + \frac{C}{r+1} + \frac{D}{r+2} \\ &&&= \frac{1}{2r^2} + \frac{1}{r+1} - \frac{1}{4(r+2)} - \frac{3}{4r} \end{align*} So \begin{align*} && \sum_{r=1}^N \frac{1}{r^2(r+1)(r+2)} &= \sum_{r=1}^N \left ( \frac{1}{2r^2} + \frac{1}{r+1} - \frac{1}{4(r+2)} - \frac{3}{4r} \right ) \\ &&&= \frac12 \sum_{r=1}^N \frac{1}{r^2} + \frac{1}{2} - \frac14 \cdot \frac1{3} - \frac34 \frac11 + \\ &&& \quad \quad \quad + \frac13 - \frac14\frac14 - \frac34\frac12 + \\ &&& \quad \quad \quad + \frac14 - \frac14\frac15 - \frac34\frac13 + \\ &&&\quad \quad \quad+ \cdots + \\ &&&\quad \quad \quad + \frac1{N+1} - \frac14\frac1{N+2} - \frac34\frac1N \\ &&&= \frac12 \sum_{r=1}^N \frac{1}{r^2} +\frac14\frac12 - \frac34\frac11-\frac14\frac1{N+2}+\frac34 \frac1{N+1} \\ \\ \Rightarrow && \sum_{r=1}^{\infty} \frac{1}{r^2(r+1)(r+2)} &= \lim_{N \to \infty} \left [ \frac12 \sum_{r=1}^N \frac{1}{r^2} +\frac14\frac12 - \frac34\frac11-\frac14\frac1{N+2}+\frac34 \frac1{N+1}\right] \\ &&&= \frac{\pi^2}{12} -\frac58 \end{align*}
  3. Notice that \(\frac{1}{r^2(r+1)^2} - \frac{2}{r^2(r+1)} = \frac{1-2(r+1)}{r^2(r+1)^2} = \frac{-1-2r}{r^2(r+1)^2} = \frac{1}{(r+1)^2} - \frac{1}{r^2}\) and so \begin{align*} && \sum_{r=1}^N \frac{1}{r^2(r+1)^2} &= \sum_{r=1}^N \left ( \frac{2}{r^2(r+1)} +\frac{1}{(r+1)^2} - \frac{1}{r^2} \right) \\ &&&= \sum_{r=1}^N \frac{2}{r^2(r+1)} +\frac{1}{(N+1)^2} - 1 \\ \end{align*} and the result follows as \(N \to \infty\)
[There is a beautiful paper by KConrad about this question: https://kconrad.math.uconn.edu/blurbs/analysis/series_acceleration.pdf]

2024 Paper 3 Q7
D: 1500.0 B: 1500.0

In this question, you need not consider issues of convergence. For positive integer \(n\) let \[\mathrm{f}(n) = \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + \ldots\] and \[\mathrm{g}(n) = \frac{1}{n+1} - \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} - \ldots\,.\]

  1. Show, by considering a geometric series, that \(0 < \mathrm{f}(n) < \dfrac{1}{n}\).
  2. Show, by comparing consecutive terms, that \(0 < \mathrm{g}(n) < \dfrac{1}{n+1}\).
  3. Show, for positive integer \(n\), that \((2n)!\,\mathrm{e} - \mathrm{f}(2n)\) and \(\dfrac{(2n)!}{\mathrm{e}} + \mathrm{g}(2n)\) are both integers.
  4. Show that if \(q\,\mathrm{e} = \dfrac{p}{\mathrm{e}}\) for some positive integers \(p\) and \(q\), then \(q\,\mathrm{f}(2n) + p\,\mathrm{g}(2n)\) is an integer for all positive integers \(n\).
  5. Hence show that the number \(\mathrm{e}^2\) is irrational.

2023 Paper 2 Q5
D: 1500.0 B: 1500.0

  1. The sequence \(x_n\) for \(n = 0, 1, 2, \ldots\) is defined by \(x_0 = 1\) and by \[x_{n+1} = \frac{x_n + 2}{x_n + 1}\] for \(n \geqslant 0\).
    1. Explain briefly why \(x_n \geqslant 1\) for all \(n\).
    2. Show that \(x_{n+1}^2 - 2\) and \(x_n^2 - 2\) have opposite sign, and that \[\left|x_{n+1}^2 - 2\right| \leqslant \tfrac{1}{4}\left|x_n^2 - 2\right|\,.\]
    3. Show that \[2 - 10^{-6} \leqslant x_{10}^2 \leqslant 2\,.\]
  2. The sequence \(y_n\) for \(n = 0, 1, 2, \ldots\) is defined by \(y_0 = 1\) and by \[y_{n+1} = \frac{y_n^2 + 2}{2y_n}\] for \(n \geqslant 0\).
    1. Show that, for \(n \geqslant 0\), \[y_{n+1} - \sqrt{2} = \frac{(y_n - \sqrt{2})^2}{2y_n}\] and deduce that \(y_n \geqslant 1\) for \(n \geqslant 0\).
    2. Show that \[y_n - \sqrt{2} \leqslant 2\left(\frac{\sqrt{2}-1}{2}\right)^{2^n}\] for \(n \geqslant 1\).
    3. Using the fact that \[\sqrt{2} - 1 < \tfrac{1}{2}\,,\] or otherwise, show that \[\sqrt{2} \leqslant y_{10} \leqslant \sqrt{2} + 10^{-600}\,.\]

2022 Paper 2 Q2
D: 1500.0 B: 1500.0

A sequence \(u_n\), where \(n = 1, 2, \ldots\), is said to have \emph{degree} \(d\) if \(u_n\), as a function of \(n\), is a polynomial of degree \(d\).

  1. Show that, in any sequence \(u_n\) \((n = 1, 2, \ldots)\) that satisfies \(u_{n+1} = \frac{1}{2}(u_{n+2} + u_n)\) for all \(n \geqslant 1\), there is a constant difference between successive terms. Deduce that any sequence \(u_n\) for which \(u_{n+1} = \frac{1}{2}(u_{n+2} + u_n)\), for all \(n \geqslant 1\), has degree at most 1.
  2. The sequence \(v_n\) \((n = 1, 2, \ldots)\) satisfies \(v_{n+1} = \frac{1}{2}(v_{n+2} + v_n) - p\) for all \(n \geqslant 1\), where \(p\) is a non-zero constant. By writing \(v_n = t_n + pn^2\), show that the sequence \(v_n\) has degree 2. Given that \(v_1 = v_2 = 0\), find \(v_n\) in terms of \(n\) and \(p\).
  3. The sequence \(w_n\) \((n = 1, 2, \ldots)\) satisfies \(w_{n+1} = \frac{1}{2}(w_{n+2} + w_n) - an - b\) for all \(n \geqslant 1\), where \(a\) and \(b\) are constants with \(a \neq 0\). Show that the sequence \(w_n\) has degree 3. Given that \(w_1 = w_2 = 0\), find \(w_n\) in terms of \(n\), \(a\) and \(b\).

2022 Paper 2 Q3
D: 1500.0 B: 1500.0

The Fibonacci numbers are defined by \(F_0 = 0\), \(F_1 = 1\) and, for \(n \geqslant 0\), \(F_{n+2} = F_{n+1} + F_n\).

  1. Prove that \(F_r \leqslant 2^{r-n} F_n\) for all \(n \geqslant 1\) and all \(r \geqslant n\).
  2. Let \(S_n = \displaystyle\sum_{r=1}^{n} \frac{F_r}{10^r}\). Show that \[\sum_{r=1}^{n} \frac{F_{r+1}}{10^{r-1}} - \sum_{r=1}^{n} \frac{F_r}{10^{r-1}} - \sum_{r=1}^{n} \frac{F_{r-1}}{10^{r-1}} = 89S_n - 10F_1 - F_0 + \frac{F_n}{10^n} + \frac{F_{n+1}}{10^{n-1}}\,.\]
  3. Show that \(\displaystyle\sum_{r=1}^{\infty} \frac{F_r}{10^r} = \frac{10}{89}\) and that \(\displaystyle\sum_{r=7}^{\infty} \frac{F_r}{10^r} < 2 \times 10^{-6}\). Hence find, with justification, the first six digits after the decimal point in the decimal expansion of \(\dfrac{1}{89}\).
  4. Find, with justification, a number of the form \(\dfrac{r}{s}\) with \(r\) and \(s\) both positive integers less than \(10000\) whose decimal expansion starts \[0.0001010203050813213455\ldots\,.\]

2021 Paper 2 Q8
D: 1500.0 B: 1500.0

  1. Show that, for \(n = 2, 3, 4, \ldots\), \[ \frac{d^2}{dt^2}\bigl[t^n(1-t)^n\bigr] = n\,t^{n-2}(1-t)^{n-2}\bigl[(n-1) - 2(2n-1)t(1-t)\bigr]. \]
  2. The sequence \(T_0, T_1, \ldots\) is defined by \[ T_n = \int_0^1 \frac{t^n(1-t)^n}{n!}\,e^t\,dt. \] Show that, for \(n \geqslant 2\), \[ T_n = T_{n-2} - 2(2n-1)T_{n-1}. \]
  3. Evaluate \(T_0\) and \(T_1\) and deduce that, for \(n \geqslant 0\), \(T_n\) can be written in the form \[ T_n = a_n + b_n e, \] where \(a_n\) and \(b_n\) are integers (which you should not attempt to evaluate).
  4. Show that \(0 < T_n < \dfrac{e}{n!}\) for \(n \geqslant 0\). Given that \(b_n\) is non-zero for all~\(n\), deduce that \(\dfrac{-a_n}{b_n}\) tends to \(e\) as \(n\) tends to infinity.

2021 Paper 3 Q8
D: 1500.0 B: 1500.0

A sequence \(x_1, x_2, \ldots\) of real numbers is defined by \(x_{n+1} = x_n^2 - 2\) for \(n \geqslant 1\) and \(x_1 = a\).

  1. Show that if \(a > 2\) then \(x_n \geqslant 2 + 4^{n-1}(a-2)\).
  2. Show also that \(x_n \to \infty\) as \(n \to \infty\) if and only if \(|a| > 2\).
  3. When \(a > 2\), a second sequence \(y_1, y_2, \ldots\) is defined by \[ y_n = \frac{Ax_1 x_2 \cdots x_n}{x_{n+1}}, \] where \(A\) is a positive constant and \(n \geqslant 1\). Prove that, for a certain value of \(a\), with \(a > 2\), which you should find in terms of \(A\), \[ y_n = \frac{\sqrt{x_{n+1}^2 - 4}}{x_{n+1}} \] for all \(n \geqslant 1\). Determine whether, for this value of \(a\), the second sequence converges.


Solution:

  1. Claim \(x_n \geqslant 2 + 4^{n-1}(a-2)\) Proof: (By induction) Base case: Note that when \(n = 1\), \(x_1 = a = 2 + 1 \cdot(a - 2)\). Inductive step, suppose true for some \(n\), then \begin{align*} && x_{n+1} &= x_n^2 - 2 \\ &&&\geq (2+4^{n-1}(a-2))^2 - 2 \\ &&&= 4 + 4^{2n-2}(a-2)^2 + 4^n(a-2) - 2 \\ &&&= 2 + 4^{n}(a-2) + 4^{2n-2}(a-2)^2 \\ &&&\geq 2 + 4^{n+1-1}(a-2) \end{align*} as required,
  2. (\(\Leftarrow\)) Suppose \(a > 2\) then \(x_n \geq 2+4^{n-1}(a-2) \to \infty\) as required. Suppose \(a < -2\) then \(x_2 > 4 -2 = 2\) so the sequence starting from \(x_2\) clearly diverges for the same reason. (\(\Rightarrow\)) suppose \(|x_n| \leq 2\) then \(x_{n+1} = x_n^2 - 2 \leq 2\) so the sequence is bounded and cannot tend to \(\infty\).
  3. Suppose \(y_n = \frac{Ax_1x_2 \cdots x_n}{x_{n+1}}\) and notice that \(x_{n+1}^2 - 4 = (x_n^2 -2)^2 - 4 = x_n^4 - 4x_n^2 = x_n^2(x_n^2-4)\). In particular, \(\frac{\sqrt{x_{n+1}^2-4}}{x_{n+1}} = \frac{x_n\sqrt{x_n^2-4}}{x_{n+1}} = \frac{x_n x_{n-1} \cdots x_1 \sqrt{x_1^2-4}}{x_{n+1}}\) Therefore if \(A = \sqrt{a^2-4}\) \(y_{n+1} = \frac{\sqrt{x_{n+1}^2-4}}{x_{n+1}}\). Notice that \begin{align*} && y_n &= \frac{\sqrt{x_{n+1}^2-4}}{x_{n+1}} \\ &&&= \sqrt{1 - \frac{4}{x_{n+1}^2}} \to 1 \end{align*}

2020 Paper 2 Q3
D: 1500.0 B: 1500.0

A sequence \(u_1, u_2, \ldots, u_n\) of positive real numbers is said to be unimodal if there is a value \(k\) such that \[u_1 \leqslant u_2 \leqslant \ldots \leqslant u_k\] and \[u_k \geqslant u_{k+1} \geqslant \ldots \geqslant u_n.\] So the sequences \(1, 2, 3, 2, 1\);\ \(1, 2, 3, 4, 5\);\ \(1, 1, 3, 3, 2\) and \(2, 2, 2, 2, 2\) are all unimodal, but \(1, 2, 1, 3, 1\) is not. A sequence \(u_1, u_2, \ldots, u_n\) of positive real numbers is said to have property \(L\) if \(u_{r-1}u_{r+1} \leqslant u_r^2\) for all \(r\) with \(2 \leqslant r \leqslant n-1\).

  1. Show that, in any sequence of positive real numbers with property \(L\), \[u_{r-1} \geqslant u_r \implies u_r \geqslant u_{r+1}.\] Prove that any sequence of positive real numbers with property \(L\) is unimodal.
  2. A sequence \(u_1, u_2, \ldots, u_n\) of real numbers satisfies \(u_r = 2\alpha u_{r-1} - \alpha^2 u_{r-2}\) for \(3 \leqslant r \leqslant n\), where \(\alpha\) is a positive real constant. Prove that, for \(2 \leqslant r \leqslant n\), \[u_r - \alpha u_{r-1} = \alpha^{r-2}(u_2 - \alpha u_1)\] and, for \(2 \leqslant r \leqslant n-1\), \[u_r^2 - u_{r-1}u_{r+1} = (u_r - \alpha u_{r-1})^2.\] Hence show that the sequence consists of positive terms and is unimodal, provided \(u_2 > \alpha u_1 > 0\). In the case \(u_1 = 1\) and \(u_2 = 2\), prove by induction that \(u_r = (2-r)\alpha^{r-1} + 2(r-1)\alpha^{r-2}\). Let \(\alpha = 1 - \dfrac{1}{N}\), where \(N\) is an integer with \(2 \leqslant N \leqslant n\). In the case \(u_1 = 1\) and \(u_2 = 2\), prove that \(u_r\) is largest when \(r = N\).

2020 Paper 3 Q8
D: 1500.0 B: 1500.0

A sequence \(u_k\), for integer \(k \geqslant 1\), is defined as follows. \[ u_1 = 1 \] \[ u_{2k} = u_k \text{ for } k \geqslant 1 \] \[ u_{2k+1} = u_k + u_{k+1} \text{ for } k \geqslant 1 \]

  1. Show that, for every pair of consecutive terms of this sequence, except the first pair, the term with odd subscript is larger than the term with even subscript.
  2. Suppose that two consecutive terms in this sequence have a common factor greater than one. Show that there are then two consecutive terms earlier in the sequence which have the same common factor. Deduce that any two consecutive terms in this sequence are co-prime (do not have a common factor greater than one).
  3. Prove that it is not possible for two positive integers to appear consecutively in the same order in two different places in the sequence.
  4. Suppose that \(a\) and \(b\) are two co-prime positive integers which do not occur consecutively in the sequence with \(b\) following \(a\). If \(a > b\), show that \(a-b\) and \(b\) are two co-prime positive integers which do not occur consecutively in the sequence with \(b\) following \(a-b\), and whose sum is smaller than \(a+b\). Find a similar result for \(a < b\).
  5. For each integer \(n \geqslant 1\), define the function \(\mathrm{f}\) from the positive integers to the positive rational numbers by \(\mathrm{f}(n) = \dfrac{u_n}{u_{n+1}}\). Show that the range of \(\mathrm{f}\) is all the positive rational numbers, and that \(\mathrm{f}\) has an inverse.