(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.