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\includegraphics[max width=\textwidth, alt={}, center]{23b06b1c-997f-425d-ae3d-bd4cc1295605-10_771_1146_260_497}
The diagram shows the curve with equation \(\mathrm { y } = \ln ( 1 + \mathrm { x } )\) for \(0 \leqslant x \leqslant 1\), together with a set of \(n\) rectangles each of width \(\frac { 1 } { n }\).
- By considering the sum of the areas of these rectangles, show that \(\int _ { 0 } ^ { 1 } \ln ( 1 + x ) d x < U _ { n }\), where
$$U _ { n } = \frac { 1 } { n } \ln \frac { ( 2 n ) ! } { n ! } - \ln n$$
- Use a similar method to find, in terms of \(n\), a lower bound \(\mathrm { L } _ { \mathrm { n } }\) for \(\int _ { 0 } ^ { 1 } \ln ( 1 + x ) \mathrm { d } x\).
- By simplifying \(\mathrm { U } _ { \mathrm { n } } - \mathrm { L } _ { \mathrm { n } }\), show that \(\lim _ { \mathrm { n } \rightarrow \infty } \left( \mathrm { U } _ { \mathrm { n } } - \mathrm { L } _ { \mathrm { n } } \right) = 0\).