The diagram shows the parts of the graph of \(y = \frac{2}{1+x}\) between \(x = 1\) and \(x = 2\). Prove that the total area of the four rectangles, each of equal width, as shown in the diagram below, may be expressed as&nbsp;\(\sum_{r=1}^{4} \frac{2}{8+r}\).[2]
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In general, when there are \(n\) rectangles, each of equal width, under the curve between \(x = 1\) and \(x = 2\) (instead of just \(4\)), find an expression, in the form \(\sum_{r=1}^{n} \mathrm{f}(r)\), for their total area.
Hence show that&nbsp;\(\sum_{r=1}^{n} \frac{2}{2n+r} &lt; 2 \ln \left(\frac{3}{2}\right)\).
Deduce that&nbsp;\(\sum_{r=1}^{n} \frac{2}{2n+r} &lt; 2 \ln \left(\frac{3}{2}\right) &lt; \sum_{r=1}^{n} \frac{2}{2n+r-1}\).

The diagram shows the parts of the graph of \(y = \frac{2}{1+x}\) between \(x = 1\) and \(x = 2\). Prove that the total area of the four rectangles, each of equal width, as shown in the diagram below, may be expressed as \(\sum_{r=1}^{4} \frac{2}{8+r}\).

In general, when there are \(n\) rectangles, each of equal width, under the curve between \(x = 1\) and \(x = 2\) (instead of just \(4\)), find an expression, in the form \(\sum_{r=1}^{n} \mathrm{f}(r)\), for their total area.

Hence show that \(\sum_{r=1}^{n} \frac{2}{2n+r} < 2 \ln \left(\frac{3}{2}\right)\).

Deduce that \(\sum_{r=1}^{n} \frac{2}{2n+r} < 2 \ln \left(\frac{3}{2}\right) < \sum_{r=1}^{n} \frac{2}{2n+r-1}\).