(a)     Using the diagram, show that $\frac{1}{{{4^3}}} + \frac{1}{{{5^3}}} + \frac{1}{{{6^3}}} + ... < \int_3^\infty  {\frac{1}{{{x^3}}}{\text{d}}x < \frac{1}{{{3^3}}} + \frac{1}{{{4^3}}} + \frac{1}{{{5^3}}} + ...}$ .

(b)     Hence find upper and lower bounds for $\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^3}}}}$.

(a)     The area under the curve is sandwiched between the sum of the areas of the lower rectangles and the upper rectangles.     M2

Therefore

$1 \times \frac{1}{{{4^3}}} + 1 \times \frac{1}{{{5^3}}} + 1 \times \frac{1}{{{6^3}}} + ... < \int_3^\infty  {\frac{{{\text{d}}x}}{{{x^3}}} < 1 \times \frac{1}{{{3^3}}} + 1 \times \frac{1}{{{4^3}}} + 1 \times \frac{1}{{{5^3}}} + ...}$     A1

which leads to the printed result.

[3 marks]

(b)     We note first that

$\int_3^\infty  {\frac{{{\text{d}}x}}{{{x^3}}} = \left[ { - \frac{1}{{2{x^2}}}} \right]_3^\infty  = \frac{1}{{18}}}$     M1A1

Consider first

$\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^3}}}}  = 1 + \frac{1}{{{2^3}}} + \frac{1}{{{3^3}}} + \left( {\frac{1}{{{4^3}}} + \frac{1}{{{5^3}}} + \frac{1}{{{6^3}}} + ...} \right)$     M1A1

$< 1 + \frac{1}{8} + \frac{1}{{27}} + \frac{1}{{18}}$     M1A1

$= \frac{{263}}{{216}}{\text{ (1.22)}}$ (which is an upper bound)     A1

$\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^3}}}}  = 1 + \frac{1}{{{2^3}}} + \left( {\frac{1}{{{3^3}}} + \frac{1}{{{4^3}}} + \frac{1}{{{5^3}}} + ...} \right)$     M1A1

$> 1 + \frac{1}{8} + \frac{1}{{18}}$     M1A1

$= \frac{{85}}{{72}}\left( {\frac{{255}}{{216}}} \right){\text{ (1.18)}}$ (which is a lower bound)     A1

[12 marks]

Total [15 marks]

Many candidates failed to give a convincing argument to establish the inequality. In (b), few candidates progressed beyond simply evaluating the integral.