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2004 Paper 3 Q7
D: 1700.0 B: 1500.0
integration reduction formula inequality logarithm summation bounding recurrence relation ln 2

For \(n=1\), \(2\), \(3\), \(\ldots\,\), let \[ I_n = \int_0^1 {t^{n-1} \over \l t+1 \r^n} \, \mathrm{d} t \, . \] By considering the greatest value taken by \(\displaystyle {t \over t+1}\) for \(0 \le t \le 1\) show that \(I_{n+1} < {1 \over 2} I_{n}\,\). Show also that \(\; \displaystyle I_{n+1}= - \frac 1{\; n\, 2^n} + I_{n}\,\). Deduce that \(\; \displaystyle I_n < \frac1 {\; n \, 2^{n-1}}\,\). Prove that \[ \ln 2 = \sum_{r=1}^n {1 \over \; r\, 2^r} + I_{n+1} \] and hence show that \({2 \over 3} < \ln 2 < {17 \over 24}\,\).

Solution: \begin{align*} && \frac{t}{t+1} &= 1 - \frac{1}{t+1} \geq \frac12 \\ \Rightarrow && I_{n+1} &= \int_0^1 \frac{t^{n}}{(t+1)^{n+1}} \d t \\ &&&= \int_0^1\frac{t}{t+1} \frac{t^{n-1}}{(t+1)^{n}} \d t \\ &&&< \int_0^1\frac12\frac{t^{n-1}}{(t+1)^{n}} \d t \\ &&&= \frac12 I_n \\ \\ && I_{n+1} &= \int_0^1 \frac{t^{n}}{(t+1)^{n+1}} \d t \\ &&&= \left [ t^n \frac{(1+t)^{-n}}{-n} \right]_0^1 +\frac1n \int_0^1 n t^{n-1}(1+t)^{-n} \d t \\ &&&= -\frac{1}{n2^n} + I_n \\ \Rightarrow && \frac12 I_n &> -\frac1{n2^n} + I_n \\ \Rightarrow && \frac{1}{n2^{n-1}} &> I_n \end{align*} \begin{align*} && \ln 2 &= \int_0^1 \frac{1}{1+t} \d t \\ &&&= I_1 \\ &&&= \frac1{2} + I_2 \\ &&&= \frac1{2} + \frac{1}{2 \cdot 2^2} + I_3 \\ &&&= \sum_{r=1}^n \frac{1}{r2^r} + I_{n+1} \\ \\ && \ln 2 &= \frac12 + \frac18 + \frac1{24} + I_4 \\ \Rightarrow && \ln 2 &> \frac12 + \frac18 + \frac1{24} = \frac{12+3+1}{24} = \frac{16}{24} = \frac23 \\ \Rightarrow && \ln 2 &= \frac12 + \frac18 + I_3 \\ &&&< \frac12 + \frac18 +\frac{1}{3 \cdot 4} \\ &&&< \frac{12}{24} + \frac{3}{24} + \frac{2}{24} = \frac{17}{24} \end{align*}

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1991 Paper 2 Q5
D: 1600.0 B: 1516.0
reduction formulae integration curve sketching tan powers series alternating series ln 2 Leibniz formula for pi

Give a rough sketch of the function \(\tan^{k}\theta\) for \(0\leqslant\theta\leqslant\frac{1}{4}\pi\) in the two cases \(k=1\) and \(k\gg1\) (i.e. \(k\) is much greater than 1). Show that for any positive integer \(n\) \[ \int_{0}^{\frac{1}{4}\pi}\tan^{2n+1}\theta\,\mathrm{d}\theta=(-1)^{n}\left(\tfrac{1}{2}\ln2+\sum_{m=1}^{n}\frac{(-1)^{m}}{2m}\right), \] and deduce that \[ \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m}=\tfrac{1}{2}\ln2. \] Show similarly that \[ \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m-1}=\frac{\pi}{4}. \]

Solution:

TikZ diagram
Let \(\displaystyle I_n = \int_0^{\pi/4} \tan^{n} \theta \, \d \theta\), then \begin{align*} I_0 &= \int_0^{\pi/4} \tan \theta \d \theta \\ &= \left [ -\ln \cos \theta \right]_0^{\pi/4} \\ &= -\ln \frac{1}{\sqrt{2}} - 0 \\ &= \frac12 \ln 2 \\ \\ \\ I_{2n+1} &= \int_0^{\pi/4} \tan^{2n+1} \theta \, \d \theta \\&= \int_0^{\pi/4} \tan^{2n-1} \theta \tan ^2 \theta \, \d \theta \\ &= \int_0^{\pi/4} \tan^{2n-1} \theta (\sec^2 \theta - 1) \, \d \theta \\ &= \int_0^{\pi/4} \tan^{2n-1} \theta \sec^2 \theta - \tan^{2n-1} \theta \, \d \theta \\ &= \left[ \frac{1}{2n} \tan^{2n} \theta \right]_0^{\pi/4} - I_{2n-1} \\ &= \frac1{2n} - I_{2n-1} \end{align*} Therefore we can see that \(\displaystyle I_{2n+1} = (-1)^{n}\left(\tfrac{1}{2}\ln2+\sum_{m=1}^{n}\frac{(-1)^{m}}{2m}\right)\). As we can see as \(n \to \infty\), \(I_n \to 0\) Therefore \begin{align*} && 0 &= \tfrac{1}{2}\ln2+\lim_{n \to \infty} \sum_{m=1}^{n}\frac{(-1)^{m}}{2m} \\ \Rightarrow && \tfrac{1}{2}\ln2 &= \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m} \end{align*} \begin{align*} && I_{-1} &= \int_0^{\pi/4} 1 \d \theta \\ &&&= \frac{\pi}{4} \end{align*} Therefore \(\displaystyle I_{2n} = (-1)^n \left ( \frac{\pi}{4} + \sum_{m=1}^n \frac{(-1)^m}{2m-1} \right)\) and since \(I_{2m} \to 0\) the same result follows.

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