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AS & A Level · Practice questions · Rectangles, bounds and sums

Squeeze the sum, find the limit.

Six original Cambridge-style questions on rectangles and the area under a curve: bounding an integral, proving inequalities for sums, bounding a numerical sum and evaluating limits of sums as definite integrals.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
The method

Sketch, choose the corner, then integrate.

01
Calculation
[3 marks]

Use four rectangles of width \(0.5\) to find a lower and an upper bound for \(\displaystyle\int_{0}^{2} \frac{1}{1 + x}\,\mathrm{d}x\). Give your answers as fractions and check that the exact value lies between them.

  • \(\dfrac{1}{1 + x}\) is decreasing; heights at \(x = 0, 0.5, 1, 1.5, 2\) are \(1, \frac{2}{3}, \frac{1}{2}, \frac{2}{5}, \frac{1}{3}\). ✓
  • Lower (right ends): \(0.5\left(\frac{2}{3} + \frac{1}{2} + \frac{2}{5} + \frac{1}{3}\right) = 0.5 \times 1.9 = \frac{19}{20}\). ✓
  • Upper (left ends): \(0.5\left(1 + \frac{2}{3} + \frac{1}{2} + \frac{2}{5}\right) = \frac{77}{60}\). ✓
  • Exact value \(\ln 3 \approx 1.0986\), between \(0.95\) and \(1.2833\). ✓
02
Proof
[4 marks]

By considering rectangles under the curve \(y = \dfrac{1}{x^{2}}\), show that \(\displaystyle\sum_{r=2}^{n} \frac{1}{r^{2}} < 1 - \frac{1}{n}\) for \(n \ge 2\). Deduce that \(\displaystyle\sum_{r=1}^{n} \frac{1}{r^{2}} < 2\) for all positive integers \(n\).

  • \(\dfrac{1}{x^{2}}\) is decreasing for \(x > 0\), so the rectangle on \([r - 1, r]\) with height \(\dfrac{1}{r^{2}}\) lies below the curve. ✓
  • Adding for \(r = 2\) to \(n\): \[ \sum_{r=2}^{n} \frac{1}{r^{2}} < \int_{1}^{n} \frac{\mathrm{d}x}{x^{2}} = \Big[-\frac{1}{x}\Big]_{1}^{n} \] which is \(1 - \dfrac{1}{n}\). ✓
  • Add the term \(r = 1\), which is 1: \(\displaystyle\sum_{r=1}^{n} \frac{1}{r^{2}} < 2 - \frac{1}{n} < 2\). ✓
  • For \(n = 1\) the sum is \(1 < 2\), so the result holds for all \(n\). ✓
03
Calculation
[5 marks]

(a) By considering rectangles above and below the curve \(y = \sqrt{x}\), show that \[ \tfrac{2}{3}n^{3/2} < \sum_{r=1}^{n} \sqrt{r} \] \[ \sum_{r=1}^{n} \sqrt{r} < \tfrac{2}{3}\left((n + 1)^{3/2} - 1\right) \]

(b) Hence find bounds for \(\displaystyle\sum_{r=1}^{100} \sqrt{r}\), correct to one decimal place in the safe direction.

  • (a) \(\sqrt{x}\) is increasing. The rectangle on \([r - 1, r]\) with height \(\sqrt{r}\) lies above the curve; adding for \(r = 1\) to \(n\) gives \(\displaystyle\sum \sqrt{r} > \int_{0}^{n} \sqrt{x}\,\mathrm{d}x = \tfrac{2}{3}n^{3/2}\). ✓
  • The rectangle on \([r, r + 1]\) with height \(\sqrt{r}\) lies below the curve; adding gives \(\displaystyle\sum \sqrt{r} < \int_{1}^{n+1} \sqrt{x}\,\mathrm{d}x = \tfrac{2}{3}\left((n + 1)^{3/2} - 1\right)\). ✓
  • (b) \(n = 100\): lower bound \(\tfrac{2}{3}(1000) = 666.66\dots\), so the sum exceeds \(666.6\). ✓
  • Upper bound \(\tfrac{2}{3}(101^{3/2} - 1) = 676.02\dots\), so the sum is less than \(676.1\). ✓
  • \(666.6 < \displaystyle\sum_{r=1}^{100}\sqrt{r} < 676.1\) (the true value is \(671.46\dots\)). ✓
04
Calculation
[3 marks]

Find the exact value of \(\displaystyle\lim_{n\to\infty} \sum_{r=1}^{n} \frac{n}{(n + 2r)^{2}}\).

  • Divide top and bottom by \(n^{2}\): \(\dfrac{n}{(n + 2r)^{2}} = \dfrac{1}{n}\cdot\dfrac{1}{\left(1 + 2\frac{r}{n}\right)^{2}}\). ✓
  • The limit is \[ \int_{0}^{1} \frac{\mathrm{d}x}{(1 + 2x)^{2}} = \Big[-\frac{1}{2(1 + 2x)}\Big]_{0}^{1} \] ✓
  • \(= -\dfrac{1}{6} + \dfrac{1}{2} = \dfrac{1}{3}\). ✓
05
Analysis
[4 marks]

Express \(\displaystyle\sum_{r=1}^{n} \frac{\sqrt{n + 3r}}{n^{3/2}}\) in the form \(\displaystyle\frac{1}{n}\sum_{r=1}^{n} f\!\left(\frac{r}{n}\right)\), and hence find the exact value of its limit as \(n \to \infty\).

  • \(\sqrt{n + 3r} = \sqrt{n}\,\sqrt{1 + 3\frac{r}{n}}\), and \(\dfrac{\sqrt{n}}{n^{3/2}} = \dfrac{1}{n}\). ✓
  • So the sum is \(\dfrac{1}{n}\displaystyle\sum_{r=1}^{n} f\!\left(\frac{r}{n}\right)\) with \(f(x) = \sqrt{1 + 3x}\). ✓
  • The limit is \[ \int_{0}^{1} (1 + 3x)^{1/2}\,\mathrm{d}x = \Big[\tfrac{2}{9}(1 + 3x)^{3/2}\Big]_{0}^{1} \] ✓
  • \(= \tfrac{2}{9}(8 - 1) = \dfrac{14}{9}\). ✓
06
Analysis
[5 marks]

Let \(\displaystyle H_{n} = \sum_{r=1}^{n} \frac{1}{r}\).

(a) Using rectangles and the curve \(y = \dfrac{1}{x}\), show that \(\ln(n + 1) < H_{n} < 1 + \ln n\) for \(n \ge 2\).

(b) Deduce that \(H_{n} < 5\) whenever \(n \le 54\), and that \(H_{n} > 5\) whenever \(n \ge 148\).

  • (a) \(\dfrac{1}{x}\) decreases. Rectangles on \([r, r + 1]\) of height \(\dfrac{1}{r}\) lie above the curve: \(H_{n} > \int_{1}^{n+1} \frac{1}{x}\,\mathrm{d}x = \ln(n + 1)\). ✓
  • Rectangles on \([r - 1, r]\) of height \(\dfrac{1}{r}\), \(r = 2\) to \(n\), lie below: \(H_{n} - 1 < \int_{1}^{n} \frac{1}{x}\,\mathrm{d}x = \ln n\), so \(H_{n} < 1 + \ln n\). ✓
  • (b) \(H_{n}\) increases with \(n\). For \(n \le 54\): \(H_{n} \le H_{54} < 1 + \ln 54 = 4.988\dots < 5\). ✓
  • For \(n \ge 148\): \(H_{n} \ge H_{148} > \ln 149 = 5.0039\dots > 5\). ✓
  • (The bounds alone cannot settle \(55 \le n \le 147\); in fact \(H_{n}\) first passes 5 at \(n = 83\).) ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.