Prove that \(\mathop {\lim }\limits_{H \to \infty } \int_a^H {\frac{1}{{{x^2}}}{\text{d}}x} \) exists and find its value in terms of \(a{\text{ (where }}a \in {\mathbb{R}^ + })\).

Use the integral test to prove that \(\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2}}}} \) converges.

Let \(\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2}}}}  = L\) .

The diagram below shows the graph of \(y = \frac{1}{{{x^2}}}\).

(i)     Shade suitable regions on a copy of the diagram above and show that

\(\sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}}  + \int_{k + 1}^\infty  {\frac{1}{{{x^2}}}} {\text{d}}x < L\) .

(ii)     Similarly shade suitable regions on another copy of the diagram above and

show that \(L < \sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}}  + \int_k^\infty  {\frac{1}{{{x^2}}}} {\text{d}}x\) .

Hence show that \(\sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}}  + \frac{1}{{k + 1}} < L < \sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}}  + \frac{1}{k}\)

You are given that \(L = \frac{{{\pi ^2}}}{6}\).

By taking k = 4 , use the upper bound and lower bound for L to find an upper bound and lower bound for \(\pi \) . Give your bounds to three significant figures.

\(\mathop {\lim }\limits_{H \to \infty } \int_a^H {\frac{1}{{{x^2}}}{\text{d}}x}  = \mathop {\lim }\limits_{H \to \infty } \left[ {\frac{{ - 1}}{x}} \right]_a^H\)     A1

\(\mathop {\lim }\limits_{H \to \infty } \left( {\frac{{ - 1}}{H} + \frac{1}{a}} \right)\)     A1

\( = \frac{1}{a}\)     A1

[3 marks]

as \(\left\{ {\frac{1}{{{n^2}}}} \right\}\) is a positive decreasing sequence we consider the function \(\frac{1}{{{x^2}}}\)

we look at \(\int_1^\infty  {\frac{1}{{{x^2}}}} {\text{d}}x\)     M1

\(\int_1^\infty  {\frac{1}{{{x^2}}}} {\text{d}}x = 1\)     A1

since this is finite (allow “limit exists” or equivalent statement)     R1

\(\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2}}}} \) converges     AG

[3 marks]

(i)

attempt to shade rectangles     M1

correct start and finish points for rectangles     A1

since the area shaded is less that the area of the required staircase we have     R1

\(\sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}}  + \int_{k + 1}^\infty  {\frac{1}{{{x^2}}}} {\text{d}}x < L\)     AG

(ii)

attempt to shade rectangles     M1

correct start and finish points for rectangles     A1

since the area shaded is greater that the area of the required staircase we have     R1

\(L < \sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}}  + \int_k^\infty  {\frac{1}{{{x^2}}}} {\text{d}}x\)     AG

Note: Alternative shading and rearranging of the inequality is acceptable.

[6 marks]

\(\int_{k + 1}^\infty  {\frac{1}{{{x^2}}}} {\text{d}}x = \frac{1}{{k + 1}},{\text{ }}\int_k^\infty  {\frac{1}{{{x^2}}}} {\text{d}}x = \frac{1}{k}\)     A1A1

\(\sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}}  + \frac{1}{{k + 1}} < L < \sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}}  + \frac{1}{k}\)     AG

[2 marks]

\(\frac{{205}}{{144}} + \frac{1}{5} < \frac{{{\pi ^2}}}{6} < \frac{{205}}{{144}} + \frac{1}{4}{\text{ }}\left( {1.6236... < \frac{{{\pi ^2}}}{6} < 1.6736...} \right)\)     A1

\(\sqrt {6\left( {\frac{{205}}{{144}} + \frac{1}{5}} \right)}  < \pi  < \sqrt {6\left( {\frac{{205}}{{144}} + \frac{1}{4}} \right)} \)     (M1)

\(3.12 < \pi  < 3.17\)     A1     N2

[3 marks]

Most candidates correctly obtained the result in part (a). Many then failed to realise that having obtained this result once it could then simply be stated when doing parts (b) and (d)

Most candidates correctly obtained the result in part (a). Many then failed to realise that having obtained this result once it could then simply be stated when doing parts (b) and (d)

In part (b) the calculation of the integral as equal to 1 only scored 2 of the 3 marks. The final mark was for stating that ‘because the value of the integral is finite (or ‘the limit exists’ or an equivalent statement) then the series converges. Quite a few candidates left out this phrase.

Most candidates correctly obtained the result in part (a). Many then failed to realise that having obtained this result once it could then simply be stated when doing parts (b) and (d)

Candidates found part (c) difficult. Very few drew the correct series of rectangles and some clearly had no idea of what was expected of them.

Most candidates correctly obtained the result in part (a). Many then failed to realise that having obtained this result once it could then simply be stated when doing parts (b) and (d)

Though part (e) could be done without doing any of the previous parts of the question many students were probably put off by the notation because only a minority attempted it.