2022 HCI J1 BT Q7

It is given that \({{u}_{r}}=\frac{1}{\left( r+2k \right)!}\), where \(r\in {{\mathbb{Z}}^{+}}\) and \(k\) is a positive constant.

1. Show that \({{u}_{r}}-{{u}_{r+1}}=\frac{r+2k}{\left( r+2k+1 \right)!}\).[1]
2. Hence use the method of differences to find \(\sum\limits_{r=1}^{n}{\frac{r+2k}{\left( r+2k+1 \right)!}}\) in terms of \(n\) and \(k\). You need not simplify your answer.[3]

Let \({{S}_{n}}\) denote the sum of the first \(n\) terms of the series

\(\frac{1}{4\,\,\,2!}+\frac{1}{5\,\,\,3!}+\frac{1}{6\,\,\,4!}+...\).

3. By finding a suitable integer value of k and using the result obtained in part (ii), find \({{S}_{n}}\) in terms of \(n\) .[2]
4. Deduce that \({{S}_{n}}<\frac{1}{6}\) for all \(n\in {{\mathbb{Z}}^{+}}\).[1]
5. Find the least value of \(n\) for which \({{S}_{n}}\) is within \({{10}^{-20}}\) of the sum to infinity.[2]