Problems

In this question, you need not consider issues of convergence. For positive integer \(n\) let \[\mathrm{f}(n) = \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + \ldots\] and \[\mathrm{g}(n) = \frac{1}{n+1} - \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} - \ldots\,.\]

1. Show, by considering a geometric series, that \(0 < \mathrm{f}(n) < \dfrac{1}{n}\).
2. Show, by comparing consecutive terms, that \(0 < \mathrm{g}(n) < \dfrac{1}{n+1}\).
3. Show, for positive integer \(n\), that \((2n)!\,\mathrm{e} - \mathrm{f}(2n)\) and \(\dfrac{(2n)!}{\mathrm{e}} + \mathrm{g}(2n)\) are both integers.
4. Show that if \(q\,\mathrm{e} = \dfrac{p}{\mathrm{e}}\) for some positive integers \(p\) and \(q\), then \(q\,\mathrm{f}(2n) + p\,\mathrm{g}(2n)\) is an integer for all positive integers \(n\).
5. Hence show that the number \(\mathrm{e}^2\) is irrational.

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*}

1. Evaluate \(b_n\).
2. Show that \(0
3. Deduce that \(a_n=n!{\rm e}-[n!{\rm e}]\) (where \([x]\) is the integer part of \(x\)).
4. Hence show that \(\mathrm{e}\) is irrational.