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2015 Paper 2 Q1
D: 1600.0 B: 1516.0
Taylor series logarithm inequality summation harmonic series calculus monotonic function series bounds

  1. By use of calculus, show that \(x- \ln(1+x)\) is positive for all positive \(x\). Use this result to show that \[ \sum_{k=1}^n \frac 1 k > \ln (n+1) \,. \]
  2. By considering \( x+\ln (1-x)\), show that \[ \sum_{k=1}^\infty \frac 1 {k^2} <1+ \ln 2 \,. \]

Solution:

  1. Consider \(f(x) = x - \ln (1+ x)\), then \(f'(x) = 1 - \frac{1}{1+x} = \frac{x}{1+x} > 0\) if \(x >0\). Therefore \(f(x)\) is strictly increasing on the positive reals. Since \(f(0) = 0\) we must have \(f(x) > 0\) for all positive \(x\), ie \(x - \ln(1+x)\) is positive for all positive \(x\). \begin{align*} \sum_{k=1}^n \frac1k &\underbrace{>}_{x > \ln(1+x)} \sum_{k=1}^n \ln \left (1 + \frac1k \right ) \\ &= \sum_{k=1}^n \ln \left ( \frac{k+1}{k} \right ) \\ &= \sum_{k=1}^n \left ( \ln (k+1) - \ln (k) \right) \\ &= \ln (n+1) - \ln 1 \\ &= \ln (n+1) \end{align*}
  2. Let \(g(x) = x + \ln (1-x)\) ,then \(g'(x) = 1 - \frac{1}{1-x} = \frac{-x}{1-x} < 0\) if \(0 < x < 1\) and \(g(0) = 0\). Therefore \(g(x)\) is decreasing and hence negative on \(0 < x < 1\), in particular \(x < -\ln(1-x) \) \begin{align*} \sum_{k=2}^n \frac1{k^2} &\underbrace{<}_{x < -\ln(1+x)} \sum_{k=2}^n - \ln \left (1-\frac1{k^2} \right) \\ &= -\sum_{k=2}^n \ln \left ( \frac{k^2-1}{k^2}\right) \\ &= \sum_{k=2}^n \l 2 \ln k - \ln(k-1) - \ln(k+1) \r \\ &= \ln n - \ln(n+1) - \ln 0+\ln 2 \\ &= \ln 2 + \ln \frac{n}{n+1} \end{align*} as \(n \to \infty\) we must have \(\displaystyle \sum_{k=2}^{\infty} \frac1{k^2} < \ln 2\) ie \[ \sum_{k=1}^\infty \frac 1 {k^2} <1+ \ln 2\]

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2009 Paper 2 Q6
D: 1600.0 B: 1516.0
sequences Fibonacci sequence recurrence relation inequality summation series bounds geometric series comparison test

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 \(\displaystyle S=\sum_{i=1}^\infty \dfrac1 {F_i}\,\).

  1. Show that \(\displaystyle \frac 1{F_i} > \frac1{2F_{i-1}}\,\) for \(i\ge4\) and deduce that \(S > 3\,\). Show also that \(S < 3\frac23\,\).
  2. 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 \(\frac{1}{F_i} > \frac1{2F_{i-1}}\)

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