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2016 Paper 2 Q8
Evaluate the integral \[ \hphantom{ \ \ \ \ \ \ \ \ \ (m> \tfrac12)\,.} \int_{m-\frac12} ^\infty \frac 1{x^2}\, \d x { \ \ \ \ \ \ \ \ \ (m > \tfrac12)\,.} \] Show by means of a sketch that \[ \sum_{r=m}^n \frac 1 {r^2} \approx \int_{m-\frac12}^{n+\frac12} \frac1 {x^2} \, \d x \,, \tag{\(*\)} \] where \(m\) and \(n\) are positive integers with \(m < n\).
- You are given that the infinite series \(\displaystyle \sum_{r=1}^\infty \frac 1 {r^2}\) converges to a value denoted by \(E\). Use \((*)\) to obtain the following approximations for \(E\): \[ E\approx 2\,; \ \ \ \ E\approx \frac53\,; \ \ \ \ E\approx \frac{33}{20} \,.\]
- Show that, when \(r\) is large, the error in approximating \(\dfrac 1{r^2}\) by \(\displaystyle \int_{r-\frac12}^{r+\frac12} \frac 1 {x^2} \, \d x\) is approximately \(\dfrac 1{4r^4}\,\). Given that \(E \approx 1.645\), show that \(\displaystyle \sum_{r=1}^\infty \frac1{r^4} \approx 1.08\, \).
Solution: \begin{align*} && \int_{m-\frac12}^\infty \frac{1}{x^2} \d x &= \lim_{K \to \infty} \left [ -x^{-1} \right]_{m-\frac12}^K \\ &&&= \frac{1}{m-\frac12} - \lim_{K \to \infty }\frac{1}K \\ &&&= \frac{1}{m-\frac12} \end{align*}
- \(\,\) \begin{align*} E &\approx \int_{1-\frac12}^\infty \frac1{x^2} \d x = \frac{1}{1-\frac12} = 2 \\ E &\approx 1 + \int_{2-\frac12}^\infty \frac1{x^2} \d x= 1 + \frac{1}{2 - \frac12} = \frac53 \\ E &\approx 1 +\frac14 + \int_{3-\frac12}^\infty \frac1{x^2} \d x= \frac54 + \frac{1}{3-\frac12} \\ &= \frac54+\frac{2}{5} = \frac{33}{20} \end{align*}
- The error is \begin{align*} && \epsilon &= \int_{r-\frac12}^{r+\frac12} \frac 1 {x^2} \, \d x - \frac1{r^2} \\ &&&= \frac{1}{r-\frac12} - \frac{1}{r + \frac12} - \frac1{r^2} \\ &&&= \frac{1}{r^2 - \frac14} - \frac1{r^2} \\ &&&= \frac{\frac14}{r^2(r^2-\frac14)} \\ &&&\approx \frac{1}{4r^4} \end{align*} Therefore \begin{align*} && \sum_{n=1}^\infty \frac1{r^4} &\approx 4 \left ( 1 +\frac14 + \int_{3-\frac12}^\infty \frac1{x^2} \d x-\sum_{r=1}^\infty \frac{1}{r^2} \right) + 1 + \frac{1}{2^4}\\ &&&= 4 \left ( \frac{33}{20}-1.645 \right) + 1 + \frac{1}{2^4} \\ &&&= 4 \left ( 1.65-1.645 \right) + 1 + \frac{1}{2^4} \\ &&&= 1.0825 \approx 1.08 \end{align*}
2009 Paper 2 Q6
The Fibonacci sequence \(F_1\), \(F_2\), \(F_3\), \(\ldots\) is defined by \(F_1=1\), \(F_2= 1\) and \[ F_{n+1} = F_n+F_{n-1} \qquad\qquad (n\ge 2). \] Write down the values of \(F_3\), \(F_4\), \(\ldots\), \(F_{10}\). Let \(\displaystyle S=\sum_{i=1}^\infty \dfrac1 {F_i}\,\).
- Show that \(\displaystyle \frac 1{F_i} > \frac1{2F_{i-1}}\,\) for \(i\ge4\) and deduce that \(S > 3\,\). Show also that \(S < 3\frac23\,\).
- Show further that \(3.2 < S < 3.5\,\).
Solution: \begin{array}{c|r} n & F_n \\ \hline 1 & 1 \\ 2 & 1 \\ 3 & 2 \\ 4 & 3 \\ 5 & 5 \\ 6 & 8 \\ 7 & 13 \\ 8 & 21 \\ 9 & 34 \\ 10 & 55 \end{array} \begin{questionparts} \item Claim: \(\frac1{F_i} > \frac1{2F_{i-1}}\) for \(i \geq 4\). Proof: Since \(F_i = F_{i-1}+F_{i-2}\) and \(F_i > 1\) for \(i \geq 1\) we have \(F_i > F_{i-1}\) for \(i \geq 3\). In particular we have \(F_i = F_{i-1}+F_{i-2} < 2F_{i-1}\) for \(i -1 \geq 3\) or \(i \geq 4\). Therefore \(\frac{1}{F_i} > \frac1{2F_{i-1}}\)
1997 Paper 3 Q7
For each positive integer \(n\), let \begin{align*} a_n&=\frac1{n+1}+\frac1{(n+1)(n+2)}+\frac1{(n+1)(n+2)(n+3)}+\cdots;\\ b_n&=\frac1{n+1}+\frac1{(n+1)^2}+\frac1{(n+1)^3}+\cdots. \end{align*}
- Evaluate \(b_n\).
- Show that \(0
- Deduce that \(a_n=n!{\rm e}-[n!{\rm e}]\) (where \([x]\) is the integer part of \(x\)).
- Hence show that \(\mathrm{e}\) is irrational.