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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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2020 Paper 2 Q11
D: 1500.0 B: 1500.0
conditional probability geometric series recurrence relation infinite series repeated trials coin tossing probability

A coin is tossed repeatedly. The probability that a head appears is \(p\) and the probability that a tail appears is \(q = 1 - p\).

  1. A and B play a game. The game ends if two successive heads appear, in which case A wins, or if two successive tails appear, in which case B wins. Show that the probability that the game never ends is \(0\). Given that the first toss is a head, show that the probability that A wins is \(\dfrac{p}{1 - pq}\). Find and simplify an expression for the probability that A wins.
  2. A and B play another game. The game ends if three successive heads appear, in which case A wins, or if three successive tails appear, in which case B wins. Show that \[\mathrm{P}(\text{A wins} \mid \text{the first toss is a head}) = p^2 + (q + pq)\,\mathrm{P}(\text{A wins} \mid \text{the first toss is a tail})\] and give a similar result for \(\mathrm{P}(\text{A wins} \mid \text{the first toss is a tail})\). Show that \[\mathrm{P}(\text{A wins}) = \frac{p^2(1-q^3)}{1-(1-p^2)(1-q^2)}.\]
  3. A and B play a third game. The game ends if \(a\) successive heads appear, in which case A wins, or if \(b\) successive tails appear, in which case B wins, where \(a\) and \(b\) are integers greater than \(1\). Find the probability that A wins this game. Verify that your result agrees with part (i) when \(a = b = 2\).

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2018 Paper 2 Q5
D: 1600.0 B: 1505.3
Taylor series binomial expansion integration logarithmic series exponential series term-by-term integration substitution infinite series definite integral Frullani integral

In this question, you should ignore issues of convergence.

  1. Write down the binomial expansion, for \(\vert x \vert<1\,\), of \(\;\dfrac{1}{1+x}\,\) and deduce that %. By considering %$ %\displaystyle \int \frac 1 {1+x} \, \d x %\,, %$ %show that \[ \displaystyle \ln (1+x) = -\sum_{n=1}^\infty \frac {(-x)^n}n \, \] for \(\vert x \vert <1 \,\).
  2. Write down the series expansion in powers of \(x\) of \(\displaystyle \e^{-ax}\,\). Use this expansion to show that \[ \int_0^\infty \frac {\left(1- \e^{-ax}\right)\e^{-x}}x \,\d x = \ln(1+a) \ \ \ \ \ \ \ (\vert a \vert <1)\,. \]
  3. Deduce the value of \[ \int_0^1 \frac{x^p - x^q}{\ln x} \, \d x \ \ \ \ \ \ (\vert p\vert <1, \ \vert q\vert <1) \,. \]

Solution:

  1. \begin{align*} && \frac1{1+x} &= 1 - x+ x^2 - x^3+ \cdots \\ \Rightarrow && \int_0^x \frac{1}{1+t} \d t &= \int_0^x \sum_{n=0}^{\infty} (-t)^n \d t \\ &&&= \left [\sum_{n=0}^{\infty} -\frac{(-t)^{n+1}}{n+1} \right]_0^x \\ \Rightarrow &&\ln(1+x)&=- \sum_{n=1}^\infty \frac{(-x)^n}{n} \end{align*}
  2. \begin{align*} && e^{-ax} &= \sum_{n=0}^\infty \frac{(-a)^n}{n!} x^n \\ \Rightarrow && \int_0^{\infty} \frac{1}{x} \left (1-e^{-ax} \right)e^{-x} \d x &= \int_0^{\infty} \frac{1}{x} \left (-\sum_{n=1}^\infty \frac{(-a)^n}{n!}x^n \right)e^{-x} \d x \\ &&&= -\int_0^{\infty} \sum_{n=1}^\infty \frac{(-a)^n}{n!} x^{n-1} e^{-x} \d x \\ &&&= -\sum_{n=1}^\infty \frac{(-a)^n}{n!} \int_0^{\infty} x^{n-1} e^{-x} \d x \\ &&&= -\sum_{n=1}^\infty \frac{(-a)^n}{n!} (n-1)! \\ &&&= -\sum_{n=1}^\infty \frac{(-a)^n}{n} \\ &&&= \ln (1+a) \end{align*}
  3. \begin{align*} && \int_0^1 \frac{x^p - x^q}{\ln x} \, \d x &= \int_0^1 \frac{x^p(1 - x^{q-p})}{\ln x} \, \d x \\ e^{-u} = x, \d x = -e^{-u} \d u: &&&=\int_{u=\infty}^{0} \frac{e^{-pu}-e^{-qu}}{-u} (-e^{-u})\d u \\ &&&= \int_0^\infty \frac{e^{-u}(e^{-qu}-e^{-pu})}{u} \d u \\ &&&= \int_0^\infty \frac{e^{-(1+q)u}(1-e^{-(p-q)u})}{u} \d u \\ v = (1+q)u, \d v = (1+q) \d u: &&&=\int_0^{\infty} \frac{e^{-v}(1-e^{-\left(\frac{p-q}{1+q}\right)v}}{v}\d v \\ &&&= \ln \left(1 + \frac{p-q}{1+q} \right) \\ &&&= \ln \left ( \frac{1+p}{1+q} \right) \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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2015 Paper 1 Q13
D: 1500.0 B: 1501.1
probability negative binomial distribution geometric distribution fair die binomial expansion conditional probability infinite series

A fair die with faces numbered \(1, \ldots, 6\) is thrown repeatedly. The events \(A\), \(B\), \(C\), \(D\) and \(E\) are defined as follows. \begin{align*} A: && \text{the first 6 arises on the \(n\)th throw.}\\ B: && \text{at least one 5 arises before the first 6.} \\ C: && \text{at least one 4 arises before the first 6.}\\ D: && \text{exactly one 5 arises before the first 6.}\\ E: && \text{exactly one 4 arises before the first 6.} \end{align*} Evaluate the following probabilities:

  1. \(\P(A)\)
  2. \(\P(B)\)
  3. \(\P(B\cap C)\)
  4. \(\P(D)\)
  5. \(\P(D\cup E)\)
For some parts of this question, you may want to make use of the binomial expansion in the form: \[ (1-x)^{-n} = 1 +nx +\frac {n(n+1)}2 x^2 + \cdots + \frac {(n+r-1)!}{r! (n-1)!}x^r +\cdots\ .\]

Solution:

  1. \(\,\) \begin{align*} \mathbb{P}(A) &= \mathbb{P}(\text{the first 6 arises on the \(n\)th throw.}) \\ &= \mathbb{P}(\text{\(n-1\) not 6s, followed by a 6.})\\ &= \left ( \frac56\right)^{n-1} \cdot \frac16 = \frac{5^{n-1}}{6^n} \end{align*}
  2. There is nothing special about \(5\) or \(6\), so which comes first is \(50:50\), therefore this probability is \(\frac12\)
  3. There is nothing special about \(4\), \(5\) or \(6\) so this is the probability that \(6\) appears last out of these three numbers, hence \(\frac13\)
  4. \(\,\) \begin{align*} \mathbb{P}(D) &= \mathbb{P}(\text{exactly one 5 arises before the first 6.}) \\ &=\sum_{n=2}^{\infty} \mathbb{P}(\text{exactly one 5 arises before the first 6 which appears on the \(n\)th roll.}) \\ &= \sum_{n=2}^{\infty} \binom{n-1}{1} \left ( \frac46 \right)^{n-2} \frac16 \cdot \frac16 \\ &= \frac1{36} \sum_{n=2}^{\infty} (n-1) \left ( \frac23 \right)^{n-2} \\ &= \frac1{36} \sum_{n=1}^{\infty} n \left ( \frac23 \right)^{n-1} \\ &= \frac1{36} \frac{1}{\left ( 1- \frac23 \right)^2} = \frac14 \end{align*}
  5. \(\,\) \begin{align*} \mathbb{P}(D \cup E) &= \mathbb{P}(D) + \mathbb{P}(E) - \mathbb{P}(D \cap E) \\ &= \frac12 - \mathbb{P}(D \cap E) \\ &=\frac12 - \sum_{n=3}^{\infty} \mathbb{P}(\text{exactly one 5 and one 4 arises before the first 6 which appears on the \(n\)th roll.}) \\ &=\frac12 - \sum_{n=3}^{\infty} 2\binom{n-1}{2} \left ( \frac36 \right)^{n-3}\cdot \frac16 \cdot \frac16 \cdot \frac16 \\ &=\frac12 - \frac2{6^3}\sum_{n=3}^{\infty} \frac{(n-1)(n-2)}{2} \left ( \frac12 \right)^{n-3} \\ &=\frac12 - \frac2{6^3}\frac{1}{(1-\tfrac12)^3}\\ &= \frac12 - \frac{2}{27} \\ &= \frac{23}{54} \end{align*}

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2015 Paper 2 Q5
D: 1600.0 B: 1484.9
trigonometry arctan proof by induction telescoping summation tan addition formula right triangle infinite series

In this question, the \(\mathrm{arctan}\) function satisfies \(0\le \arctan x <\frac12 \pi\) for \(x\ge0\,\).

  1. Let \[ S_n= \sum_{m=1}^n \arctan \left(\frac1 {2m^2}\right) \,, \] for \(n=1, 2, 3, \ldots\) . Prove by induction that \[ \tan S_n = \frac n {n+1} \,. \] Prove also that \[ S_n = \arctan \frac n {n+1} \,. \]
  2. In a triangle \(ABC\), the lengths of the sides \(AB\) and \(BC\) are \(4n^2\) and \(4n^4-1\), respectively, and the angle at \(B\) is a right angle. Let \(\angle BCA = 2\alpha_n\). Show that \[ \sum_{n=1}^\infty \alpha_n = \tfrac14\pi \,. \]

Solution:

  1. Claim: \(\tan S_n = \frac n {n+1}\) Proof: (By Induction) Base case: (\(n=1\)): \begin{align*} && \tan \left ( \sum_{m=1}^1 \arctan \left ( \frac{1}{2m^2} \right) \right) &= \tan \left ( \arctan \left ( \frac{1}{2} \right) \right) \\ &&&= \frac12 = \frac{1}{1+1} \end{align*} Therefore the base case is true. Inductive step: Suppose our statement is true for some \(n = k\), ie \begin{align*} && \frac{k}{k+1} &= \tan \left ( \sum_{m=1}^k \arctan \left ( \frac{1}{2m^2} \right) \right) \\ \Rightarrow && \tan S_{k+1} &= \tan \left ( \sum_{m=1}^k \arctan \left ( \frac{1}{2m^2} \right) + \arctan \left ( \frac{1}{2 (k+1)^2} \right) \right) \\ &&&= \frac{\tan S_k + \tan \left ( \arctan \left ( \frac{1}{2 (k+1)^2} \right) \right)}{1-\tan S_k \tan \left ( \arctan \left ( \frac{1}{2 (k+1)^2} \right) \right)} \\ &&&= \frac{\frac{k}{k+1} + \frac{1}{2(k+1)^2}}{1-\frac{k}{k+1} \frac{1}{2(k+1)^2}} \\ &&&= \frac{2k(k+1)^2+(k+1)}{2(k+1)^3-k} \\ &&&= \frac{k+1}{(k+1)+1} \end{align*} Therefore it is true for \(n=k+1\). Conclusion: Therefore by the principle of mathematical induction since our statement is true for \(n=1\) and if it is true for \(n=k\) it is true for \(n=k+1\) it is true for all \(n\geq1\) Since \(S_n < \frac12 \pi\) for all \(n\), we must have \(\arctan \frac{n}{n+1} = S_n\)
  2. \(\tan (2\alpha_n) = \frac{4n^2}{4n^4-1} = \frac{2n^2+2n^2}{(2n^2)(2n^2)-1} = \frac{\frac{1}{2n^2}+\frac{1}{2n^2}}{1-\frac{1}{2n^2}\frac{1}{2n^2}} \Rightarrow \tan (\alpha_n) = \arctan \frac{1}{2n^2}\). In particular \(\displaystyle \sum_{n=1}^{N} \alpha_n = \arctan \frac{n}{n+1} \Rightarrow \sum_{n=1}^{\infty} \alpha_n \to \arctan 1 = \frac{\pi}{4} \)

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2010 Paper 2 Q3
D: 1600.0 B: 1500.0
sequences Fibonacci sequence recurrence relation golden ratio simultaneous equations geometric series infinite series

The first four terms of a sequence are given by \(F_0=0\), \(F_1=1\), \(F_2=1\) and \(F_3=2\). The general term is given by \[ F_n= a\lambda^n+b\mu^n\,, \tag{\(*\)} \] where \(a\), \(b\), \(\lambda\) and \(\mu\) are independent of \(n\), and \(a\) is positive.

  1. Show that \(\lambda^2 +\lambda\mu+ \mu^2 = 2\), and find the values of \(\lambda\), \(\mu\), \(a\) and \(b\).
  2. Use \((*)\) to evaluate \(F_6\).
  3. Evaluate \(\displaystyle \sum_{n=0}^\infty \frac{F_n}{2^{n+1}}\,.\)

Solution:

  1. \(\,\) \begin{align*} && 0 &= a+b \tag{1}\\ && 1 &= a\lambda -a\mu \tag{2} \\ && 1 &= a\lambda^2 -a\mu^2 \tag{3} \\ && 2 &= a\lambda^3 - a\mu^3 \tag{4} \\ (4) \div (2): && 2 & = \lambda^2+\lambda \mu + \mu^2 \\ (3) \div (2): && 1 &= \lambda + \mu \\ \Rightarrow && 2 &= \lambda^2 + \lambda(1-\lambda) + (1-\lambda)^2 \\ &&&= \lambda^2-\lambda+1\\ \Rightarrow && \lambda, \mu &= \frac{1 \pm \sqrt{5}}{2} \\ \Rightarrow && a &= \frac{1}{\lambda - \mu} = \frac{1}{\sqrt{5}} \\ \Rightarrow && b &= -\frac{1}{\sqrt{5}} \end{align*} (NB: This is Binet's formula)
  2. \(\,\) \begin{align*} F_6 &= \frac{1}{\sqrt{5}} \left ( \left ( \frac{1 +\sqrt{5}}{2} \right)^6- \left ( \frac{1 -\sqrt{5}}{2} \right)^6 \right) \\ &= \frac{1}{2^6 \sqrt{5}} \left ( (1+\sqrt{5})^6-(1-\sqrt{5})^6 \right) \\ &= \frac{1}{2^5 \cdot \sqrt{5}} \left (6 \sqrt{5} +\binom{6}{3} (\sqrt{5})^3+\binom{6}{5}(\sqrt{5})^5 \right)\\ &= \frac{1}{2^5} \left (6 +20\cdot 5+6\cdot 5^2 \right)\\ &= \frac{1}{2^5} 256 = 2^3 = 8 \end{align*} (way more painful than just computing it by adding terms!)
  3. \(\,\) \begin{align*} && \sum_{n=0}^{\infty} \frac{F_n}{2^{n+1}} &= \sum_{n=0}^{\infty} \frac{a\lambda^n + b\mu^n}{2^{n+1}} \\ &&&= \frac12 \left ( \frac{a}{1-\frac{\lambda}2} + \frac{b}{1-\frac{\mu}2} \right) \\ &&&= \frac12 \left ( \frac{2a}{2-\lambda} + \frac{2b}{2-\mu}\right) \\ &&&= \frac{2a}{4-2\lambda} + \frac{2b}{4-2\mu}\\ &&&= \frac{2a}{4-(1+\sqrt{5})} - \frac{2a}{4-(1-\sqrt{5})} \\ &&&= \frac{2}{3\sqrt{5}-5} - \frac{2}{3\sqrt{5}+5} \\ &&&= \frac{6\sqrt{5}+10-6\sqrt{5}+10}{45-25} \\ &&&= 1 \end{align*}

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2010 Paper 3 Q12
D: 1700.0 B: 1500.0
geometric distribution infinite series arithmetico-geometric series expected value probability alternating turns proof summing series

The infinite series \(S\) is given by \[ S = 1 + (1 + d)r + (1 + 2d)r^2 + \cdots + (1+nd)r^n +\cdots\; ,\] for \(\vert r \vert <1\,\). By considering \(S - rS\), or otherwise, prove that \[ S = \frac 1{1-r} + \frac {rd}{(1-r)^2} \,.\] Arthur and Boadicea shoot arrows at a target. The probability that an arrow shot by Arthur hits the target is \(a\); the probability that an arrow shot by Boadicea hits the target is \(b\). Each shot is independent of all others. Prove that the expected number of shots it takes Arthur to hit the target is \(1/a\). Arthur and Boadicea now have a contest. They take alternate shots, with Arthur going first. The winner is the one who hits the target first. The probability that Arthur wins the contest is \(\alpha\) and the probability that Boadicea wins is \(\beta\). Show that \[ \alpha = \frac a {1-a'b'}\,, \] where \(a' = 1-a\) and \(b'=1-b\), and find \(\beta\). Show that the expected number of shots in the contest is \(\displaystyle \frac \alpha a + \frac \beta b\,.\)

Solution: Notice that \begin{align*} && S - rS &= 1 + dr + dr^2 + \cdots \\ &&&= 1 + dr(1 + r+r^2+ \cdots) \\ &&&= 1 + \frac{rd}{1-r} \\ \Rightarrow && S &= \frac{1}{1-r} + \frac{rd}{(1-r)^2} \end{align*} The number of shots Arthur takes is \(\textrm{Geo}(a)\), so it's expectation is \(1/a\). The probability Arthur wins is: \begin{align*} \alpha &= a + a'b'a + (a'b')^2a + \cdots \\ &= a(1+a'b' + \cdots) \\ &= \frac{a}{1-a'b'} \\ \\ \beta &= a'b + a'b'a'b + \cdots \\ &= a'b(1+b'a' + (b'a')^2 + \cdots ) \\ &= \frac{a'b}{1-a'b'} \end{align*} The expected number of shots in the contest is: \begin{align*} E &= a + 2a'b + 3a'b'a + 4a'b'a'b + \cdots \\ &= a(1 + 3a'b' + 5(a'b')^2 + \cdots) + 2a'b(1 + 2(a'b') + 3(a'b')^2 + \cdots) \\ &= a \left ( \frac{1}{1-a'b'} + \frac{2a'b'}{(1-a'b')^2} \right) + 2a'b \left ( \frac{1}{1-a'b'} + \frac{a'b'}{(1-a'b')^2}\right) \\ &= \frac{a}{1-a'b'} \left (1 + \frac{2a'b'}{(1-a'b')} \right) + 2\frac{a'b}{1-a'b'} \left ( 1 + \frac{a'b'}{(1-a'b')}\right) \\ &= \alpha \frac{1+a'b'}{1-a'b'} + \beta \frac{2}{1-a'b'} \\ &= \alpha \frac{1+1-a-b+ab}{1-a'b'} + \beta \frac{2}{1-a'b'} \\ \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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1999 Paper 3 Q8
D: 1700.0 B: 1516.0
second order differential equation SHM piecewise function continuity conditions initial value problem infinite series integration trigonometric

The function \(y(x)\) is defined for \(x\ge0\) and satisfies the conditions \[ y=0 \mbox{ \ \ and \ \ } \frac{\d y}{\d x}=1 \mbox{ \ \ at \(x=0\)}. \] When \(x\) is in the range \(2(n-1)\pi< x <2n\pi\), where \(n\) is a positive integer, \(y(t)\) satisfies the differential equation $$ {\d^2y \over \d x^2} + n^2 y=0. $$ Both \(y\) and \(\displaystyle \frac{\d y}{\d x} \) are continuous at \(x=2n\pi\) for \(n=0,\; 1,\;2,\; \ldots\;\).

  1. Find \(y(x)\) for \(0\le x \le 2\pi\).
  2. Show that \(y(x) = \frac12 \sin 2x \) for \(2\pi\le x\le 4\pi\), and find \(y(x)\) for all \(x\ge0\).
  3. Show that $$ \int_0^\infty y^2 \,\d x = \pi \sum_{n=1}^\infty {1\over n^2} \,. $$

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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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1994 Paper 3 Q12
D: 1700.0 B: 1473.3
probability independent events geometric series tennis conditional probability algebraic manipulation infinite series

In certain forms of Tennis two players \(A\) and \(B\) serve alternate games. Player \(A\) has probability \(p\low_{A}\) of winning a game in which she serves and \(p\low_{B}\) of winning a game in which player \(B\) serves. Player \(B\) has probability \(q\low_{B}=1-p\low_{B}\) of winning a game in which she serves and probability \(q\low_{A}=1-p\low_{A}\) of winning a game in which player \(A\) serves. In Shortened Tennis the first player to lead by 2 games wins the match. Find the probability \(P_{\text{short}}\) that \(A\) wins a Shortened Tennis match in which she serves first and show that it is the same as if \(B\) serves first. In Standard Tennis the first player to lead by 2 or more games after 4 or more games have been played wins the match. Show that the probability that the match is decided in 4 games is \[ p^{2}_Ap_{B}^{2}+q_{A}^{2}q_{B}^{2}+2(p\low_{A}p\low_{B}+q\low_{A}q\low_{B})(p\low_{A}q\low_{B}+q\low_{A}p\low_{B}). \] If \(p\low_{A}=p\low_{B}=p\) and \(q\low_{A}=q\low_{B}=q,\) find the probability \(P_{\text{stan}}\) that \(A\) wins a Standard Tennis match in which she serves first. Show that \[ P_{\text{stan}}-P_{\text{short}}=\frac{p^{2}q^{2}(p-q)}{p^{2}+q^{2}}. \]

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1993 Paper 1 Q2
D: 1500.0 B: 1516.0
geometric series sum of geometric series convergence infinite series pattern recognition partial fractions limiting behaviour

If \(\left|r\right|\neq1,\) show that \[ 1+r^{2}+r^{4}+\cdots+r^{2n}=\frac{1-r^{2n+2}}{1-r^{2}}\,. \] If \(r\neq1,\) find an expression for \(\mathrm{S}_{n}(r),\) where \[ \mathrm{S}_{n}(r)=r+r^{2}+r^{4}+r^{5}+r^{7}+r^{8}+r^{10}+\cdots+r^{3n-1}. \] Show that, if \(\left|r\right|<1,\) then, as \(n\rightarrow\infty,\) \[ \mathrm{S}_{n}(r)\rightarrow\frac{1}{1-r}-\frac{1}{1-r^{3}}\,. \] If \(\left|r\right|\neq1,\) find an expression for \(\mathrm{T}_{n}(r),\) where \[ \mathrm{T}_{n}(r)=1+r^{2}+r^{3}+r^{4}+r^{6}+r^{8}+r^{9}+r^{10}+r^{12}+r^{14}+r^{15}+r^{16}+\cdots+r^{6n}. \] If \(\left|r\right|<1,\) find the limit of \(\mathrm{T}_{n}(r)\) as \(n\rightarrow\infty.\) What happens to \(\mathrm{T}_{n}(r)\) as \(n\rightarrow\infty\) in the three cases \(r>1,r=1\) and \(r=-1\)? In each case give reasons for your answer.

Solution: \begin{align*} && S &= 1 + r^2 + r^4 + \cdots + r^{2n} \\ && r^2S &= \quad \,\,\,\, r^2 + r^4 + \cdots+r^{2n}+r^{2n+2} \\ \Rightarrow && (1-r^2)S &= 1 - r^{2n+2} \\ \Rightarrow && S &= \frac{1-r^{2n+2}}{1-r^2} \end{align*} \begin{align*} && S_n(r) &= r + r^2 + r^4 + r^5 + r^7 + \cdots + r^{3n-1} \\ &&&= 1 + r + r^2 + \cdots + r^{3n} - (1 + r^3 + r^6 + r^{3n}) \\ &&&= \frac{1-r^{3n+1}}{1-r} - \frac{1-r^{3n+3}}{1-r^3} \\ \\ \Rightarrow && \lim_{n \to \infty} S_n(r) &= \frac{1-0}{1-r} - \frac{1-0}{1-r^3} = \frac{1}{1-r} - \frac{1}{1-r^3} \end{align*} \begin{align*} && T_n(r) &= 1 + r^2 + r^3 + r^4 + r^6 + \cdots + r^{6n} \\ &&&= \frac{1-r^{6n+6}}{1-r^6} + \frac{r^2-r^{6n+2}}{1-r^6} + \frac{r^3-r^{6n+3}}{1-r^6} + \frac{r^4-r^{6n+4}}{1-r^6} \\ &&&= \frac{1+r^2+r^3+r^4-r^{6n}(r^2+r^3+r^4+r^6))}{1-r^6} \\ \\ &&\lim_{n \to \infty} T_n(r) &= \frac{1+r^2+r^3+r^4}{1-r^6} \end{align*} If \(r > 1\) clear it diverges. if \(r = 1\) same story. if \(r = -1\) the sums in blocks of \(4\) are all \(1+1-1+1 = 2 > 0\) and so it also diverges.

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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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1987 Paper 1 Q4
D: 1500.0 B: 1500.0
sequences infinite series geometric series logarithms change of base convergence summation

Show that the sum of the infinite series \[ \log_{2}\mathrm{e}-\log_{4}\mathrm{e}+\log_{16}\mathrm{e}-\ldots+(-1)^{n}\log_{2^{2^{n}}}\mathrm{e}+\ldots \] is \[ \frac{1}{\ln(2\sqrt{2})}. \] {[}\(\log_{a}b=c\) is equivalent to \(a^{c}=b\).{]}

Solution: Let \(S = \log_{2}\mathrm{e}-\log_{4}\mathrm{e}+\log_{16}\mathrm{e}-\ldots+(-1)^{n}\log_{2^{2^{n}}}\mathrm{e}+\ldots\) then \begin{align*} S &= \sum_{n=0}^{\infty} (-1)^n \log_{2^{2^n}} e \\ &= \sum_{n=0}^{\infty} (-1)^n \frac{\log e}{\log {2^{2^n}}} \\ &= \sum_{n=0}^{\infty} (-1)^n \frac{\log e}{2^n\log {2}} \\ &= \frac{\log e}{\log 2} \sum_{n=0}^{\infty} \frac{(-1)^n}{2^n} \\ &= \frac{1}{\log_e 2} \frac{1}{1+\frac12} \\ &= \frac{1}{\ln (2^{3/2})} \\ &= \frac{1}{\ln (2 \sqrt{2})} \end{align*}

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