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2024 Paper 3 Q1
D: 1500.0 B: 1500.0
sequences and series partial fractions telescoping series algebraic identity infinite series summation

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]

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2017 Paper 3 Q1
D: 1700.0 B: 1516.0
sequences series binomial coefficients telescoping series inequality summation proof partial fractions convergence bounding sums

  1. Prove that, for any positive integers \(n\) and \(r\), \[ \frac{1}{^{n+r}\C_{r+1}} =\frac{r+1}{r} \left(\frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}}\right). \] Hence determine \[ \sum_{n=1}^{\infty}{\frac{1}{^{n+r}\C_{r+1}}} \,, \] and deduce that \ \(\displaystyle \sum_{n=2}^\infty \frac 1 {^{n+2}\C_3} = \frac12\,\).
  2. Show that, for \(n \ge 3\,\), \[ \frac{3!}{n^3} < \frac{1}{^{n+1}\C_{3}} \ \ \ \ \ \text{and} \ \ \ \ \ \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} < \frac{5!}{n^3} \,. \] By summing these inequalities for \(n \ge 3\,\), show that \[ \frac{115}{96} < \sum_{n=1}^{\infty}{\frac{1}{n^3}} < \frac{116}{96} \, . \]
{\bf Note: } \(^n\C_r\) is another notation for \(\displaystyle \binom n r \).

Solution: \begin{align*} \frac{r+1}{r} \left(\frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}}\right) &= \frac{r+1}{r} \l \frac{r!(n-1)!}{(n+r-1)!} - \frac{r!n!}{(n+r)!} \r \\ &= \frac{(r+1)!(n-1)!}{r(n+r-1)!} \l 1 - \frac{n}{n+r} \r \\ &= \frac{(r+1)!(n-1)!}{r(n+r-1)!} \frac{r}{n+r} \\ &= \frac{(r+1)!n!}{(n+r)!} \\ &= \frac{1}{^{n+r}\C_{r+1}} \end{align*} \begin{align*} \sum_{n=1}^{\infty}{\frac{1}{^{n+r}\C_{r+1}}} &= \sum_{n=1}^{\infty} \l \frac{r+1}{r} \left(\frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}}\right) \r \\ &= \frac{r+1}{r} \sum_{n=1}^{\infty} \l \frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}} \r \\ &= \frac{r+1}{r} \lim_{N \to \infty} \sum_{n=1}^{N} \l \frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}} \r \\ &= \frac{r+1}{r} \lim_{N \to \infty} \l \frac{1}{^{1+r-1}\C_{r}} - \frac{1}{^{N+r}\C_{r}}\r \\ &= \frac{r+1}{r} \frac{1}{^{1+r-1}\C_{r}} \tag{since \(\frac{1}{^{N+r}\C_{r}} \to 0\)} \\ &= \frac{r+1}{r} \end{align*} When \(r = 2\), we have: \begin{align*} && \frac{3}{2} &= \sum_{n=1}^{\infty}{\frac{1}{^{n+2}\C_{3}}} \\ && &=\frac{1}{^{1+2}\C_{3}} + \sum_{n=2}^{\infty}{\frac{1}{^{n+2}\C_{3}}} \\ && &= 1 + \sum_{n=2}^{\infty}{\frac{1}{^{n+2}\C_{3}}} \\ \Rightarrow && \sum_{n=2}^{\infty}{\frac{1}{^{n+2}\C_{3}}} &= \frac12 \end{align*} \begin{align*} \frac{1}{^{n+1}\C_{3}} &= \frac{3!}{(n+1)n(n-1)} \\ &= \frac{3!}{n^3-n} \\ &> \frac{3!}{n^3} \end{align*} \begin{align*} \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} &= \frac{5!}{(n+1)n(n-1)} - \frac{5!}{(n+2)(n+1)n(n-1)(n-2)} \\ &= \frac{5!}{n^3} \frac{n^2}{n^2-1}\l 1- \frac{1}{n^2-4} \r \\ &= \frac{5!}{n^3} \frac{n^2}{n^2-1}\l \frac{n^2-5}{n^2-4} \r \\ &= \frac{5!}{n^3} \frac{n^2(n^2-5)}{(n^2-1)(n^2-4)} \\ &< \frac{5!}{n^3} \end{align*} Since \(k(k-5) < (k-1)(k-4) \Leftrightarrow 0 < 4\), this only makes sense if \(n \geq 3\) \begin{align*} &&\frac{3!}{n^3} &< \frac{1}{^{n+1}\C_{3}} \tag{if \(n \geq 3\)} \\ \Rightarrow &&\sum_{n=3}^\infty \frac{3!}{n^3} &< \sum_{n=3}^\infty \frac{1}{^{n+1}\C_{3}} \\ \Rightarrow && \frac{6}{1^3} + \frac{6}{2^3} + \sum_{n=3}^\infty \frac{3!}{n^3} &< \frac{6}{1^3} + \frac{6}{2^3} + \sum_{n=3}^\infty \frac{1}{^{n+1}\C_{3}} \\ \Rightarrow && \sum_{n=1}^\infty \frac{3!}{n^3} &< 6 + \frac{3}{4} + \sum_{n=2}^\infty \frac{1}{^{n+2}\C_{2+1}} \\ \Rightarrow && \sum_{n=1}^\infty \frac{3!}{n^3} &< 6 + \frac{3}{4} + \frac{1}{2} = \frac{29}{4} \\ \Rightarrow && \sum_{n=1}^\infty \frac{1}{n^3} &< \frac{29}{24} = \frac{116}{96} \\ \end{align*} \begin{align*} && \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} &< \frac{5!}{n^3} \\ \Rightarrow && \sum_{n=3}^\infty \l \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} \r &< \sum_{n=3}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=3}^\infty \frac{20}{^{n+1}\C_3} - \sum_{n=3}^\infty \frac{1}{^{n+2}\C_{5}} &< \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=3}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=2}^\infty \frac{20}{^{n+2}\C_{2+1}} - \sum_{n=1}^\infty \frac{1}{^{n+4}\C_{4+1}} &< \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=3}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{120}{1^3} + \frac{120}{2^3} + \frac{20}{2} - \frac{4+1}{4} &< \sum_{n=1}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{115}{96} &< \sum_{n=1}^\infty \frac{1}{n^3} \\ \end{align*}

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2016 Paper 3 Q4
D: 1700.0 B: 1484.0
sequences telescoping series partial fractions convergence hyperbolic functions sech cosech infinite series substitution

  1. By considering \(\displaystyle \frac1{1+ x^r} - \frac1{1+ x^{r +1}}\) for \(\vert x \vert \ne 1\), simplify \[ \sum_{r=1}^N \frac{x^r}{(1+x^r)(1+x^{r+1})} \] Show that, for \(\vert x \vert <1\), \[ \sum_{r=1}^\infty \frac{x^r}{(1+x^r)(1+x^{r+1})} = \frac x {1-x^2} \]
  2. Deduce that \[ \sum_{r=1}^\infty \textrm{sech}(ry)\textrm{sech}((r + 1)y) = 2\e^{-y} \textrm{cosech}(2 y) \] for \(y > 0\). Hence simplify \[ \sum_{r=-\infty}^\infty \textrm{sech}(ry) \textrm{sech}((r + 1)y) \] for \(y>0\).

Solution:

  1. \(\,\) \begin{align*} && \frac{1}{1+x^r} - \frac{1}{1+x^{r+1}} &= \frac{1+x^{r+1}-1-x^r}{(1+x^r)(1+x^{r+1})} \\ &&&= \frac{x^r(x-1)}{(1+x^r)(1+x^{r+1})} \\ \\ && \sum_{r=1}^N \frac{x^r}{(1+x^r)(1+x^{r+1})} &= \sum_{r=1}^N \frac{1}{x-1} \left ( \frac{1}{1+x^r} - \frac{1}{1+x^{r+1}}\right) \\ &&&= \frac{1}{x-1} \Bigg ( \frac{1}{1+x} + \cdots \\ &&& \qquad \qquad \quad - \frac{1}{1+x^2} + \frac{1}{1+x^2} + \cdots \\ &&& \qquad \qquad \quad - \frac{1}{1+x^3} + \frac{1}{1+x^3} + \cdots \\ &&& \qquad \qquad \quad - \cdots \\ &&& \qquad \qquad \quad - \frac{1}{1+x^{N+1}} \Bigg ) \\ &&&= \frac{1}{x-1} \left (\frac{1}{1+x} - \frac{1}{1+x^{N+1}} \right) \\ \\ && \sum_{r=1}^{\infty} \frac{x^r}{(1+x^r)(1+x^{r+1})} &= \lim_{N\to \infty} \frac{1}{x-1} \left (\frac{1}{1+x} - \frac{1}{1+x^{N+1}} \right) \\ &&&= \frac{1}{x-1} \left ( \frac{1}{1+x} - 1\right) \\ &&&= \frac{1}{x-1} \left ( \frac{-x}{1+x} \right) \\ &&&= \frac{x}{1-x^2} \end{align*}
  2. \(\,\) \begin{align*} && \sum_{r=1}^\infty \textrm{sech}(ry)\textrm{sech}((r + 1)y) &= \sum_{r=1}^\infty \frac{4}{(e^{ry}+e^{-ry})(e^{(r+1)y}+e^{-(r+1)y})} \\ &&&=\sum_{r=1}^\infty \frac{4e^{-(2r+1)y}}{(1+e^{-2ry})(1+e^{-2(r+1)y})} \\ x = e^{-2y}: &&&= \frac{4e^{-y}e^{-2y}}{1-e^{-4y}} \\ &&&= \frac{4e^{-y}e^{-2y}}{e^{-2y}(e^{2y}-e^{-2y})} \\ &&&=2e^{-y}\textrm{cosech}(2y) \end{align*} \begin{align*} && \sum_{r=-\infty}^\infty \textrm{sech}(ry) \textrm{sech}((r + 1)y) &= \sum_{r=1}^\infty \textrm{sech}(ry) \textrm{sech}((r + 1)y) + \sum_{r=-\infty}^0 \textrm{sech}(ry) \textrm{sech}((r + 1)y) \\ &&&= 2e^{-y}\textrm{cosech}(2y) + \sum_{r=0}^\infty \textrm{sech}(-ry) \textrm{sech}(-(r-1)y) \\ &&&= 2e^{-y}\textrm{cosech}(2y) + \sum_{r=0}^\infty \textrm{sech}((r-1)y) \textrm{sech}(ry) \\ &&&= 4e^{-y}\textrm{cosech}(2y) + \textrm{sech}(y) + \textrm{sech}(-y) \\ &&&= 4e^{-y}\textrm{cosech}(2y)+2\textrm{sech}(y) \\ &&&= 4e^{-y} \frac12 \textrm{sech}(y) \textrm{cosech}(y) + 2 \textrm{sech}(y) \\ &&&= 2\textrm{sech}(y) \left ( e^{-y} \textrm{cosech}(y)+1 \right) \\ &&&= 2\textrm{sech}(y) \left ( \frac{2}{e^{2y}-1} + 1 \right) \\ &&&= 2\textrm{sech}(y) \left ( \frac{e^{2y}+1}{e^{2y}-1} \right) \\ &&&= 2 \textrm{cosech}(y) \end{align*}

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2012 Paper 3 Q8
D: 1700.0 B: 1500.0
sequences recurrence relation Fibonacci sequence proof by induction arctan telescoping series trigonometric identity

The sequence \(F_0\), \(F_1\), \(F_2\), \(\ldots\,\) is defined by \(F_0=0\), \(F_1=1\) and, for \(n\ge0\), \[ F_{n+2} = F_{n+1} + F_n \,. \]

  1. Show that \(F_0F_3-F_1F_2 = F_2F_5- F_3F_4\,\).
  2. Find the values of \(F_nF_{n+3} - F_{n+1}F_{n+2}\) in the two cases that arise.
  3. Prove that, for \(r=1\), \(2\), \(3\), \(\ldots\,\), \[ \arctan \left( \frac 1{F_{2r}}\right) =\arctan \left( \frac 1{F_{2r+1}}\right)+ \arctan \left( \frac 1{F_{2r+2}}\right) \] and hence evaluate the following sum (which you may assume converges): \[ \sum_{r=1}^\infty \arctan \left( \frac 1{F_{2r+1}}\right) \,. \]

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2003 Paper 3 Q6
D: 1700.0 B: 1516.0
trigonometric identity telescoping series summation product-to-sum formula cosine series solving trigonometric equations

Show that \[ 2\sin \frac12 \theta \, \cos r\theta = \sin\big(r+\frac12\big)\theta - \sin\big(r-\frac12\big)\theta \;. \] Hence, or otherwise, find all solutions of the equation \[ \cos a\theta + \cos (a + 1) \theta + \dots + \cos(b-2)\theta+\cos (b - 1 ) \theta = 0 \;, \] where \(a\) and \(b\) are positive integers with \(a < b-1\,\).

Solution: \begin{align*} && \sin\left(r+\frac12\right)\theta - \sin\left(r-\frac12\right)\theta &= \sin r \theta \cos \tfrac12 \theta+\cos r \theta \sin \tfrac12 \theta- \left (\sin r \theta \cos \tfrac12 \theta-\cos r \theta \sin \tfrac12 \theta \right)\\ &&&= 2 \cos r\theta \sin \tfrac12 \theta \end{align*} \begin{align*} && S &= \cos a\theta + \cos (a + 1) \theta + \dots + \cos(b-2)\theta+\cos (b - 1 ) \theta \\ && 2\sin\tfrac12 \theta S &= \sum_{r=a}^{b-1} 2\sin\tfrac12 \theta \cos r \theta \\ &&&= \sum_{r=a}^{b-1} \left ( \sin\left(r+\frac12\right)\theta - \sin\left(r-\frac12\right)\theta \right) \\ &&&= \sin \left (b-\frac12 \right)\theta - \sin \left (a -\frac12 \right)\theta \\ \Rightarrow && \sin \left (b-\frac12 \right)\theta &= \sin \left (a -\frac12 \right)\theta \\ \end{align*} Case 1: \(A = B + 2n\pi\) \begin{align*} && \left (b-\frac12 \right)\theta &= \left (a -\frac12 \right)\theta + 2n\pi \\ \Rightarrow && (b-a) \theta &= 2n \pi \\ \Rightarrow && \theta &= \frac{2n\pi}{b-a} \end{align*} Case 2: \(A = (2n+1)\pi - B\) \begin{align*} && \left (b-\frac12 \right)\theta &= (2n+1)\pi -\left (a -\frac12 \right)\theta \\ \Rightarrow && (b+a-1) \theta &= (2n+1) \pi \\ \Rightarrow && \theta &= \frac{2n\pi}{b+a-1} \end{align*}

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2002 Paper 3 Q2
D: 1700.0 B: 1500.0
trigonometry arctan addition formula proof by induction telescoping series infinite series summation inverse trigonometric functions

Prove that \(\displaystyle \arctan a + \arctan b = \arctan \l {a + b \over 1-ab} \r\,\) when \(0 < a < 1\) and \(0 < b < 1\,\). Prove by induction that, for \(n \ge 1\,\), \[ \sum_{r = 1}^n \arctan \l {1 \over r^2 + r + 1} \r = \arctan \l {n \over n+2} \r \] and hence find \[ \sum_{r = 1}^\infty \arctan \l {1 \over r^2 + r + 1} \r\,. \] Hence prove that \[ \sum_{r = 1}^\infty \arctan \l {1 \over r^2 - r + 1} \r = {\pi \over 2}\,. \]

Solution: \begin{align*} && \arctan a &\in (0, \tfrac{\pi}{4}) \\ && \arctan b &\in (0, \tfrac{\pi}{4}) \\ \Rightarrow && \arctan a+\arctan b &\in (0, \tfrac{\pi}{2}) \\ && \tan \left ( \arctan a+\arctan b \right) &= \frac{\tan \arctan a + \tan \arctan b}{1 - \tan \arctan a \tan \arctan b} \\ &&&= \frac{a+b}{1-ab} \in (0, \infty) \\ \Rightarrow && \arctan \left ( \frac{a+b}{1-ab} \right) &\in (0, \tfrac{\pi}{2}) \\ \Rightarrow && \arctan a + \arctan b &= \arctan \left ( \frac{a+b}{1-ab} \right) \end{align*} Claim: \(\displaystyle \sum_{r = 1}^n \arctan \l {1 \over r^2 + r + 1} \r = \arctan \l {n \over n+2} \r\) Proof: (By Induction): Base case (\(n=1\)): \begin{align*} && LHS &= \sum_{r=1}^1 \arctan \left ( \frac{1}{r^2+r+1} \right) \\ &&&= \arctan \left ( \frac{1}{3} \right) \\ && RHS &= \arctan \left ( \frac{1}{1+2} \right)\\ &&&= \arctan \left ( \frac{1}{3} \right) = LHS \end{align*} Inductive step, suppose true for \(n = k\), ie \begin{align*} && \sum_{r = 1}^k \arctan \l {1 \over r^2 + r + 1} \r &= \arctan \l {k \over k+2} \r \\ \Rightarrow && \sum_{r = 1}^{k+1} \arctan \l {1 \over r^2 + r + 1} \r &= \sum_{r = 1}^k \arctan \l {1 \over r^2 + r + 1} \r+ \arctan \left ( \frac{1}{(k+1)^2+(k+1)+1} \right) \\ &&&= \arctan \l {k \over k+2} \r+\arctan \left ( \frac{1}{(k+1)^2+(k+1)+1} \right) \\ &&&= \arctan \left ( \frac{{k \over k+2}+\frac{1}{(k+1)^2+(k+1)+1} }{1-\frac{k}{k+2}\frac{1}{(k+1)^2+(k+1)+1} } \right) \\ &&&= \arctan \left ( \frac{k((k+1)^2+k+1+k)+(k+2) }{(k+2)((k+1)^2+(k+1)+1)-k} \right) \\ &&&= \arctan \left ( \frac{k^3+3k^2+4k+2 }{k^3+5k^2+8k+6} \right) \\ &&&= \arctan \left ( \frac{(k+1)(k^2+2k+2) }{(k+3)(k^2+2k+2)} \right) \\ &&&= \arctan \left ( \frac{k+1 }{(k+1)+2} \right) \\ \end{align*} Therefore it is true for \(n = k+1\), therefore it is true for all \(n \geq 1\) by the principle of mathematical induction. \begin{align*} && S &= \lim_{n \to \infty} \sum_{r = 1}^n \arctan \l {1 \over r^2 + r + 1} \r \\ &&&= \lim_{n \to \infty} \arctan \l \frac{n}{n+2} \r \\ &&&= \lim_{n \to \infty} \arctan \l \frac{1}{1+2/n} \r \\ &&&=\arctan\l \lim_{n \to \infty} \frac{1}{1+2/n} \r \\ &&&= \frac{\pi}{4} \end{align*} \begin{align*} && \sum_{r = 1}^\infty \arctan \l {1 \over r^2 - r + 1} \r &= \sum_{r = 0}^\infty \arctan \left( \frac{1}{ (r+1)^2 - (r+1) + 1} \right) \\ &&&= \sum_{r = 0}^\infty \arctan \left( \frac{1}{ r^2+r+1} \right) \\ &&&= \arctan \l \frac{1}{0^2+0+1} \r + \frac{\pi}{4} \\ &&&= \frac{\pi}{2} \end{align*}

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1998 Paper 2 Q3
D: 1600.0 B: 1516.0
sequences series partial fractions telescoping series convergence recurrence relation summation product telescoping

Show that the sum \(S_N\) of the first \(N\) terms of the series $$\frac{1}{1\cdot2\cdot3}+\frac{3}{\cdot3\cdot4}+\frac{5}{3\cdot4\cdot5}+\cdots +\frac{2n-1}{n(n+1)(n+2)}+\cdots$$ is $${1\over2}\left({3\over2}+{1\over N+1}-{5\over N+2}\right).$$ What is the limit of \(S_N\) as \(N\to\infty\)? The numbers \(a_n\) are such that $$\frac{a_n}{a_{n-1}}=\frac{(n-1)(2n-1)}{(n+2)(2n-3)}.$$ Find an expression for \(a_n/a_1\) and hence, or otherwise, evaluate \(\sum\limits_{n=1}^\infty a_n\) when \(\displaystyle a_1=\frac{2}{9}\;\).

Solution: First notice by partial fractions: \begin{align*} \frac{2n-1}{n(n+1)(n+2)} &= \frac{-1/2}{n} + \frac{3}{n+1} + \frac{-5/2}{n+2} \\ &= \frac{-1}{2n} + \frac{3}{n+1} - \frac{5}{2(n+2)} \end{align*} And therefore: \begin{align*} \sum_{n = 1}^N \frac{2n-1}{n(n+1)(n+2)} &= -\frac12 \sum_{n=1}^N \frac1n +3\sum_{n=1}^N \frac1{n+1} -\frac52 \sum_{n=1}^N \frac1{n+2} \\ &= -\frac12-\frac14 + \frac{3}{2}+ \sum_{n=3}^N (3-\frac12 -\frac52)\frac1n + \frac{3}{N+1} - \frac{5}{2(N+1)} - \frac{5}{2(N+2)} \\ &= \frac12 \l \frac32+\frac1{N+1}-\frac{5}{N+2} \r \end{align*} As \(N \to \infty, S_N \to \frac{3}{4}\). \begin{align*} && \frac{a_n}{a_{n-1}}&=\frac{(n-1)(2n-1)}{(n+2)(2n-3)} \\ \Rightarrow && \frac{a_n}{a_1} &= \frac{a_n}{a_{n-1}} \cdot \frac{a_{n-1}}{a_{n-2}} \cdots \frac{a_2}{a_1} \\ &&&= \frac{(n-1)(2n-1)}{(n+2)(2n-3)} \cdot \frac{(n-2)(2n-3)}{(n+1)(2n-5)} \cdots \frac{(1)(3)}{(4)(1)} \\ &&&= \frac{(2n-1)3\cdot 2\cdot 1}{(n+2)(n+1)n} \\ &&& = \frac{6(2n-1)}{n(n+1)(n+2)} \end{align*} Therefore \(a_n = \frac{4}{3} \frac{2n-1}{n(n+1)(n+2)}\) and so our sequence is \(\frac43\) the earlier sum, ie \(1\)

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1994 Paper 2 Q6
D: 1600.0 B: 1500.0
trigonometry proof by induction telescoping series infinite series cotangent identity tan half-angle limits

Prove by induction, or otherwise, that, if \(0<\theta<\pi\), \[ \frac{1}{2}\tan\frac{\theta}{2}+\frac{1}{2^{2}}\tan\frac{\theta}{2^{2}}+\cdots+\frac{1}{2^{n}}\tan\frac{\theta}{2^{n}}=\frac{1}{2^{n}}\cot\frac{\theta}{2^{n}}-\cot\theta. \] Deduce that \[ \sum_{r=1}^{\infty}\frac{1}{2^{r}}\tan\frac{\theta}{2^{r}}=\frac{1}{\theta}-\cot\theta. \]

Solution: Claim: \(\displaystyle \sum_{r=1}^n \frac1{2^r} \tan \tfrac{\theta}{2^r} = \frac1{2^n}\cot \tfrac{\theta}{2^n} - \cot \theta\) Proof: (By Induction) Base case: \(n = 1\) \begin{align*} && LHS &= \sum_{r=1}^1 \frac1{2^r} \tan \frac{\theta}{2^r} \\ &&&= \frac1{2} \tan \frac{\theta}{2}\\ \\ && RHS &= \frac12 \cot \frac{\theta}{2} - \cot \theta \\ &&&= \frac12 \frac{1}{\tan \frac{\theta}{2}} - \frac{1-\tan^2 \frac{\theta}{2}}{2 \tan \frac{\theta}{2}} \\ &&&= \frac{1}{2} \tan \frac{\theta}{2} = LHS \end{align*} Therefore our base case is true. Assume our statement is true for some \(n=k\), then consider \(n = k+1\), ie \begin{align*} \sum_{r=1}^{k+1} \frac1{2^r} \tan \tfrac{\theta}{2^r} &= \sum_{r=1}^{k} \frac1{2^r} \tan \tfrac{\theta}{2^r} + \frac1{2^{k+1}} \tan \frac{\theta}{2^{k+1}} \\ &= \frac{1}{2^k} \cot \frac{\theta}{2^k} - \cot \theta + \frac{1}{2^{k+1}}\tan \frac{\theta}{2^{k+1}} \\ &= \frac{1}{2^{k+1}} \left (2 \cot \frac{\theta}{2^k} +\tan \frac{\theta}{2^{k+1}} \right) - \cot \theta \\ &= \frac{1}{2^{k+1}} \left (2\frac{1-\tan^2 \frac{\theta}{2^{k+1}}}{2 \tan \frac{\theta}{2^{k+1}}} + \tan \frac{\theta}{2^{k+1}} \right) - \cot \theta \\ &= \frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta \\ \end{align*} Therefore, since as \(x \to 0, x\cot x \to 1\) or \(x \cot \theta x \to \frac{1}{\theta}\) \begin{align*} \sum_{r=1}^{\infty}\frac{1}{2^{r}}\tan\frac{\theta}{2^{r}} &= \lim_{k\to \infty} \sum_{r=1}^{k}\frac{1}{2^{r}}\tan\frac{\theta}{2^{r}} \\ &= \lim_{k\to \infty} \left ( \frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta \right) \\ &= \lim_{k\to \infty}\frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta \\ &= \lim_{k\to \infty}\frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta \\ &= \lim_{k\to \infty}\frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta \\ &= \frac{1}{\theta} - \cot \theta \end{align*}

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1993 Paper 3 Q4
D: 1700.0 B: 1500.0
sequences infinite series summation power series partial fractions geometric series logarithmic series telescoping series differentiation of series

Sum the following infinite series.

  1. \[ 1 + \frac13 \bigg({\frac12}\bigg)^2 +\frac15\bigg(\frac12\bigg)^4 + \cdots + \frac{1}{2n+1} \bigg(\frac12\bigg)^{2n} + \cdots \] .
  2. \[ 2 -x -x^3 +2x^4 - \cdots + 2x^{4k} - x^{4k+1} - x^{4k+3} +\cdots \] where \(|x| < 1\).
  3. \[ \sum _{r=2}^\infty {r\, 2^{r-2} \over 3^{r-1} } \].
  4. \[ \sum_{r=2}^\infty {2 \over r(r^2-1) } \].

Solution:

  1. \begin{align*} && \sum_{i=0}^{\infty} x^{2i+1}&= \frac{x}{1-x^2} \\ \Rightarrow &&&=\frac12 \left ( \frac{1}{1-x} - \frac{1}{1+x} \right) \\ \underbrace{\Rightarrow}_{\int} && \sum_{i=0}^{\infty} \frac{1}{2i+1} x^{2i+2} &= \frac12 \left ( -\ln (1-x) - \ln(1+x) \right) \\ \underbrace{\Rightarrow}_{x = 1/2} && \sum_{i=0}^\infty \frac{1}{2i+1} \left (\frac12 \right)^{2i+2} &= -\frac12 \ln \frac12 - \frac12 \ln \frac32 \\ &&&= -\frac12 \ln \frac34 \\ &&\frac14\sum_{i=0}^\infty \frac{1}{2i+1} \left (\frac12 \right)^{2i}&= \frac12 \ln \frac43 \\ \Rightarrow&& S &= 2 \ln \frac43 \end{align*}
  2. \begin{align*} \sum_{k=0}^{\infty} \left (2x^{4k} - x^{4k+1} - x^{4k+3} \right) &= \sum_{k=0}^{\infty} \left (2 - x^{1} - x^{3} \right) x^{4k} \\ &= \frac{2-x-x^3}{1-x^4} \\ &= \frac{(1-x)(2+x+x^2)}{(1-x)(1+x+x^2+x^3)} \\ &= \frac{2+x+x^2}{1+x+x^2+x^3} \end{align*}
  3. \begin{align*} && \frac{1}{(1-x)^2} &= \sum_{r=0}^{\infty} r x^{r-1} \\ \Rightarrow && 9 &= \sum_{r=1}^{\infty} r \left ( \frac23 \right)^{r-1} \\ \Rightarrow && \sum_{r=2}^{\infty} r \left ( \frac{2^{r-2}}{3^{r-1}} \right) &= \frac12 \left ( 9 - 1 \right) \\ &&&= 4 \end{align*}
  4. \begin{align*} && \frac{2}{r(r^2-1)} &= \frac{1}{r-1} - \frac{2}{r} + \frac{1}{r+1} \\ \Rightarrow && \sum_{r=2}^{\infty} \left ( \frac{1}{r-1} - \frac{2}{r} + \frac{1}{r+1} \right) &= \sum_{r=2}^{\infty} \left ( \frac{1}{r-1} - \frac{1}{r} - \frac{1}{r} + \frac{1}{r+1} \right) \\ &&&= \sum_{r=2}^{\infty} \left ( \frac{1}{r-1} - \frac{1}{r} \right)-\sum_{r=2}^{\infty} \left ( \frac{1}{r} - \frac{1}{r+1} \right) \\ &&&= 1 - \frac12 \\ &&&= \frac12 \end{align*}

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