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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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