Question
The diagram shows part of the graph of \(y = \frac{1}{{{x^3}}}\) together with line segments parallel to the coordinate axes.
(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 \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^3}}}} \).
▶️Answer/Explanation
Markscheme
(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]
Examiners report
Many candidates failed to give a convincing argument to establish the inequality. In (b), few candidates progressed beyond simply evaluating the integral.
Question
a.Using the integral test, show that \(\sum\limits_{n = 1}^\infty {\frac{1}{{4{n^2} + 1}}} \) is convergent.[4]
b.(i) Show, by means of a diagram, that \(\sum\limits_{n = 1}^\infty {\frac{1}{{4{n^2} + 1}}} < \frac{1}{{4 \times {1^2} + 1}} + \int_1^\infty {\frac{1}{{4{x^2} + 1}}{\text{d}}x} \).
(ii) Hence find an upper bound for \(\sum\limits_{n = 1}^\infty {\frac{1}{{4{n^2} + 1}}} \)[4]
▶️Answer/Explanation
Markscheme
\(\int {\frac{1}{{4{x^2} + 1}}{\text{d}}x = \frac{1}{2}\arctan 2x + k} \) (M1)(A1)
Note: Do not penalize the absence of “+k”.
\(\int_1^\infty {\frac{1}{{4{x^2} + 1}}{\text{d}}x = \frac{1}{2}\mathop {\lim }\limits_{a \to \infty } } [\arctan 2x]_1^a\) (M1)
Note: Accept \(\frac{1}{2}[\arctan 2x]_1^\infty \).
\( = \frac{1}{2}\left( {\frac{\pi }{2} – \arctan 2} \right)\,\,\,\,\,( = 0.232)\) A1
hence the series converges AG
[4 marks]
(i)
A2
The shaded rectangles lie within the area below the graph so that \(\sum\limits_{n = 2}^\infty {\frac{1}{{4{n^2} + 1}}} < \int_1^\infty {\frac{1}{{4{x^2} + 1}}{\text{d}}x} \). Adding the first term in the series, \(\frac{1}{{4 \times {1^2} + 1}}\), gives \(\sum\limits_{n = 1}^\infty {\frac{1}{{4{n^2} + 1}}} < \frac{1}{{4 \times {1^2} + 1}} + \int_1^\infty {\frac{1}{{4{x^2} + 1}}{\text{d}}x} \) R1AG
(ii) upper bound \( = \frac{1}{5} + \frac{1}{2}\left( {\frac{\pi }{2} – \arctan 2} \right)\,\,\,\,\,( = 0.432)\) A1
[4 marks]
Examiners report
This proved to be a hard question for most candidates. A number of fully correct answers to (a) were seen, but a significant number were unable to integrate \({\frac{1}{{4{x^2} + 1}}}\) successfully. Part (b) was found the hardest by candidates with most candidates unable to draw a relevant diagram, without which the proof of the inequality was virtually impossible.
This proved to be a hard question for most candidates. A number of fully correct answers to (a) were seen, but a significant number were unable to integrate \({\frac{1}{{4{x^2} + 1}}}\) successfully. Part (b) was found the hardest by candidates with most candidates unable to draw a relevant diagram, without which the proof of the inequality was virtually impossible.
Question
a.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 \(a{\text{ (where }}a \in {\mathbb{R}^ + })\).[3]
b.Use the integral test to prove that \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \) converges.[3]
c.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 \(L < \sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}} + \int_k^\infty {\frac{1}{{{x^2}}}} {\text{d}}x\) .[6]
d.Hence show that \(\sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}} + \frac{1}{{k + 1}} < L < \sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}} + \frac{1}{k}\)[2]
e.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.[3]
▶️Answer/Explanation
Markscheme
\(\mathop {\lim }\limits_{H \to \infty } \int_a^H {\frac{1}{{{x^2}}}{\text{d}}x} = \mathop {\lim }\limits_{H \to \infty } \left[ {\frac{{ – 1}}{x}} \right]_a^H\) 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]
Examiners report
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.
Question
Consider the function \(f(x) = \frac{1}{{1 + {x^2}}},{\text{ }}x \in \mathbb{R}\).
a.Illustrate graphically the inequality, \(\frac{1}{5}\sum\limits_{r = 1}^5 {f\left( {\frac{r}{5}} \right) < \int_0^1 {f(x){\text{d}}x < \frac{1}{5}\sum\limits_{r = 0}^4 {f\left( {\frac{r}{5}} \right)} } } \).[3]
b.Use the inequality in part (a) to find a lower and upper bound for \(\pi \).[5]
c.Show that \(\sum\limits_{r = 0}^{n – 1} {{{( – 1)}^r}{x^{2r}} = \frac{{1 + {{( – 1)}^{n – 1}}{x^{2n}}}}{{1 + {x^2}}}} \).[2]
d.Hence show that \(\pi = 4\left( {\sum\limits_{r = 0}^{n – 1} {\frac{{{{( – 1)}^r}}}{{2r + 1}} – {{( – 1)}^{n – 1}}\int_0^1 {\frac{{{x^{2n}}}}{{1 + {x^2}}}{\text{d}}x} } } \right)\).[4]
▶️Answer/Explanation
Markscheme
A1A1A1
A1 for upper rectangles, A1 for lower rectangles, A1 for curve in between with \(0 \le x \le 1\)
hence \(\frac{1}{5}\sum\limits_{r = 1}^5 {f\left( {\frac{r}{5}} \right) < \int_0^1 {f(x){\text{d}}x < \frac{1}{5}\sum\limits_{r = 0}^4 {f\left( {\frac{r}{5}} \right)} } } \) AG
[3 marks]
attempting to integrate from \(0\) to \(1\) (M1)
\(\int_0^1 {f(x){\text{d}}x = [\arctan x]_0^1} \)
\( = \frac{\pi }{4}\) A1
attempt to evaluate either summation (M1)
\(\frac{1}{5}\sum\limits_{r = 1}^5 {f\left( {\frac{r}{5}} \right) < \frac{\pi }{4} < \frac{1}{5}\sum\limits_{r = 0}^4 {f\left( {\frac{r}{5}} \right)} } \)
hence \(\frac{4}{5}\sum\limits_{r = 1}^5 {f\left( {\frac{r}{5}} \right) < \pi < \frac{4}{5}\sum\limits_{r = 0}^4 {f\left( {\frac{r}{5}} \right)} } \)
so \(2.93 < \pi < 3.33\) A1A1
Note: Accept any answers that round to \(2.9\) and \(3.3\).
[5 marks]
EITHER
recognise \(\sum\limits_{r = 0}^{n – 1} {{{( – 1)}^r}{x^{2r}}} \) as a geometric series with \(r = – {x^2}\) M1
sum of \(n\) terms is \(\frac{{1 – {{( – {x^2})}^n}}}{{1 – – {x^2}}} = \frac{{1 + {{( – 1)}^{n – 1}}{x^{2n}}}}{{1 + {x^2}}}\) M1AG
OR
\(\sum\limits_{r = 0}^{n – 1} {{{( – 1)}^r}(1 + {x^2}){x^{2r}} = (1 + {x^2}){x^0} – (1 + {x^2}){x^2} + (1 + {x^2}){x^4} + \ldots } \)
\( + {( – 1)^{n – 1}}(1 + {x^2}){x^{2n – 2}}\) M1
cancelling out middle terms M1
\( = 1 + {( – 1)^{n – 1}}{x^{2n}}\) AG
[2 marks]
\(\sum\limits_{r = 0}^{n – 1} {{{( – 1)}^r}{x^{2r}} = \frac{1}{{1 + {x^2}}} + {{( – 1)}^{n – 1}}\frac{{{x^{2n}}}}{{1 + {x^2}}}} \)
integrating from \(0\) to \(1\) M1
\(\left[ {\sum\limits_{r = 0}^{n – 1} {{{( – 1)}^r}\frac{{{x^{2r + 1}}}}{{2r + 1}}} } \right]_0^1 = \int_0^1 {f(x){\text{d}}x + {{( – 1)}^{n – 1}}\int_0^1 {\frac{{{x^{2n}}}}{{1 + {x^2}}}{\text{d}}x} } \) A1A1
\(\int_0^1 {f(x){\text{d}}x = \frac{\pi }{4}} \) A1
so \(\pi = 4\left( {\sum\limits_{r = 0}^{n – 1} {\frac{{{{( – 1)}^r}}}{{2r + 1}} – {{( – 1)}^{n – 1}}\int_0^1 {\frac{{{x^{2n}}}}{{1 + {x^2}}}{\text{d}}x} } } \right)\) AG
[4 marks]
Total [14 marks]
Examiners report
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Question
a.Find the value of \(\int\limits_4^\infty {\frac{1}{{{x^3}}}{\text{d}}x} \).[3]
b.Illustrate graphically the inequality \(\sum\limits_{n = 5}^\infty {\frac{1}{{{n^3}}}} < \int\limits_4^\infty {\frac{1}{{{x^3}}}{\text{d}}x} < \sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}} \).[4]
c.Hence write down a lower bound for \(\sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}} \).[1]
d.Find an upper bound for \(\sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}} \).[3]
▶️Answer/Explanation
Markscheme
\(\int\limits_4^\infty {\frac{1}{{{x^3}}}{\text{d}}x} = \mathop {{\text{lim}}}\limits_{R \to \infty } \int\limits_4^R {\frac{1}{{{x^3}}}{\text{d}}x} \) (A1)
Note: The above A1 for using a limit can be awarded at any stage.
Condone the use of \(\mathop {{\text{lim}}}\limits_{x \to \infty } \).
Do not award this mark to candidates who use \(\infty \) as the upper limit throughout.
= \(\mathop {{\text{lim}}}\limits_{R \to \infty } \left[ { – \frac{1}{2}{x^{ – 2}}} \right]_4^R\left( { = \left[ { – \frac{1}{2}{x^{ – 2}}} \right]_4^\infty } \right)\) M1
\( = \mathop {{\text{lim}}}\limits_{R \to \infty } \left( { – \frac{1}{2}\left( {{R^{ – 2}} – {4^{ – 2}}} \right)} \right)\)
\( = \frac{1}{{32}}\) A1
[3 marks]
A1A1A1A1
A1 for the curve
A1 for rectangles starting at \(x = 4\)
A1 for at least three upper rectangles
A1 for at least three lower rectangles
Note: Award A0A1 for two upper rectangles and two lower rectangles.
sum of areas of the lower rectangles < the area under the curve < the sum of the areas of the upper rectangles so
\(\sum\limits_{n = 5}^\infty {\frac{1}{{{n^3}}}} < \int\limits_4^\infty {\frac{1}{{{x^3}}}{\text{d}}x} < \sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}} \) AG
[4 marks]
a lower bound is \(\frac{1}{{32}}\) A1
Note: Allow FT from part (a).
[1 mark]
METHOD 1
\(\sum\limits_{n = 5}^\infty {\frac{1}{{{n^3}}}} < \frac{1}{{32}}\) (M1)
\(\frac{1}{{64}} + \sum\limits_{n = 5}^\infty {\frac{1}{{{n^3}}}} = \frac{1}{{32}} + \frac{1}{{64}}\) (M1)
\(\sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}} < \frac{3}{{64}}\), an upper bound A1
Note: Allow FT from part (a).
METHOD 2
changing the lower limit in the inequality in part (b) gives
\(\sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}} < \int\limits_3^\infty {\frac{1}{{{x^3}}}{\text{d}}x} \left( { < \sum\limits_{n = 3}^\infty {\frac{1}{{{n^3}}}} } \right)\) (A1)
\(\sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}} < \mathop {{\text{lim}}}\limits_{R \to \infty } \left[ { – \frac{1}{2}{x^{ – 2}}} \right]_3^R\) (M1)
\(\sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}} < \frac{1}{{18}}\), an upper bound A1
Note: Condone candidates who do not use a limit.
[3 marks]
Examiners report
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