Year: 2016
Paper: 3
Question Number: 4
Course: UFM Pure
Section: Sequences and series, recurrence and convergence
A substantially larger number of candidates took the paper this year: 14% more than in 2015. However, the mean score was virtually identical to that in 2015. Five questions were very popular, with two being attempted by in excess of 90% of the candidates, but once again, all questions were attempted by significant numbers, with only one dipping under 10% attempting it, and every question was answered perfectly by at least one candidate. Most candidates kept to six sensible attempts, although some did several more scoring weakly overall, except in six outstanding cases that earned very high marks.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
\begin{questionparts}
\item
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 $\vert x \vert <1$,
\[
\sum_{r=1}^\infty \frac{x^r}{(1+x^r)(1+x^{r+1})} = \frac x {1-x^2}
\]
\item
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 $y>0$.
\end{questionparts}\begin{questionparts}
\item $\,$
\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*}
\item $\,$ \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*}
\end{questionparts}
Very slightly more popular than question 2 with four fifths attempting it, they did so with slightly less success. The first part of the question, being well signposted, was pretty well attempted, although there was some very poor notation for limit arguments as N was commonly taken to equal infinity. In part (ii), it was quite common for candidates to write the expression for sech(ry) in terms of positive powers of exponentials which made attempts to apply (i) invalid. Few candidates fully simplified the final result, and prior to that, many did not handle the positive and negative parts of the sum correctly with some just doubling the sum from 1 to infinity.