## Question

The function \(f(x) = \frac{{1 + ax}}{{1 + bx}}\) can be expanded as a power series in *x*, within its radius of convergence R, in the form \(f(x) \equiv 1 + \sum\limits_{n = 1}^\infty {{c_n}{x^n}} \) .

(a) (i) Show that \({c_n} = {( – b)^{n – 1}}(a – b)\).

(ii) State the value of R.

(b) Determine the values of *a* and *b* for which the expansion of *f*(*x*) agrees with that of \({{\text{e}}^x}\) up to and including the term in \({x^2}\) .

(c) Hence find a rational approximation to \({{\text{e}}^{\frac{1}{3}}}\) .

**Answer/Explanation**

## Markscheme

(a) (i) \(f(x) = (1 + ax){(1 + bx)^{ – 1}}\)

\( = (1 + ax)(1 – bx + …{( – 1)^n}{b^n}{x^n} + …\) *M1A1*

it follows that

\({c_n} = {( – 1)^n}{b^n} + {( – 1)^{n – 1}}a{b^{n – 1}}\) *M1A1*

\( = {( – b)^{n – 1}}(a – b)\) *AG*

* *

(ii) \({\text{R}} = \frac{1}{{\left| b \right|}}\) *A1*

*[5 marks]*

* *

(b) to agree up to quadratic terms requires

\(1 = – b + a,{\text{ }}\frac{1}{2} = {b^2} – ab\) *M1A1A1*

from which \(a = – b = \frac{1}{2}\) *A1*

*[4 marks]*

* *

(c) \({{\text{e}}^x} \approx \frac{{1 + 0.5x}}{{1 – 0.5x}}\) *A1*

putting \(x = \frac{1}{3}\) *M1*

\({{\text{e}}^{\frac{1}{3}}} \approx \frac{{\left( {1 + \frac{1}{6}} \right)}}{{\left( {1 – \frac{1}{6}} \right)}} = \frac{7}{5}\) *A1*

*[3 marks]*

*Total [12 marks]*

## Examiners report

Most candidates failed to realize that the first step was to write *f*(*x*) as \((1 + ax){(1 + bx)^{ – 1}}\) . Given the displayed answer to part(a), many candidates successfully tackled part(b). Few understood the meaning of the ‘hence’ in part(c).

## Question

Find the radius of convergence of the infinite series

\[\frac{1}{2}x + \frac{{1 \times 3}}{{2 \times 5}}{x^2} + \frac{{1 \times 3 \times 5}}{{2 \times 5 \times 8}}{x^3} + \frac{{1 \times 3 \times 5 \times 7}}{{2 \times 5 \times 8 \times 11}}{x^4} + \ldots {\text{ .}}\]

Determine whether the series \(\sum\limits_{n = 1}^\infty {\sin \left( {\frac{1}{n} + n\pi } \right)} \) is convergent or divergent.

**Answer/Explanation**

## Markscheme

the *n*th term is

\({u_n} = \frac{{1 \times 3 \times 5 \ldots (2n – 1)}}{{2 \times 5 \times 8 \ldots (3n – 1)}}{x^n}\) *M1A1*

(using the ratio test to test for absolute convergence)

\(\frac{{\left| {{u_{n + 1}}} \right|}}{{\left| {{u_n}} \right|}} = \frac{{(2n + 1)}}{{(3n + 2)}}\left| x \right|\) *M1A1*

\(\mathop {\lim }\limits_{n \to \infty } \frac{{\left| {{u_{n + 1}}} \right|}}{{\left| {{u_n}} \right|}} = \frac{{2\left| x \right|}}{3}\) *A1*

let *R* denote the radius of convergence

then \(\frac{{2R}}{3} = 1\) so \(r = \frac{3}{2}\) *M1A1*

**Note:** Do not penalise the absence of absolute value signs.

*[7 marks]*

using the compound angle formula or a graphical method the series can be written in the form *(M1)*

\(\sum\limits_{n = 1}^\infty {{u_n}} \) where \({u_n} = {( – 1)^n}\sin \left( {\frac{1}{n}} \right)\) *A2*

since \(\frac{1}{n} < \frac{\pi }{2}\) *i.e.* an angle in the first quadrant, *R1*

it is an alternating series *R1*

\({u_n} \to 0{\text{ as }}n \to \infty \) *R1*

and \(\left| {{u_{n + 1}}} \right| < \left| {{u_n}} \right|\) *R1*

it follows that the series is convergent *R1*

*[8 marks]*

## Examiners report

Solutions to this question were generally disappointing. In (a), many candidates were unable even to find an expression for the *n*th term so that they could not apply the ratio test.

Solutions to this question were generally disappointing. In (b), few candidates were able to rewrite the *n*th term in the form \(\sum {{{( – 1)}^n}\sin \left( {\frac{1}{n}} \right)} \) so that most candidates failed to realise that the series was alternating.

## Question

Consider the power series \(\sum\limits_{k = 1}^\infty {k{{\left( {\frac{x}{2}} \right)}^k}} \).

(i) Find the radius of convergence.

(ii) Find the interval of convergence.

Consider the infinite series \(\sum\limits_{k = 1}^\infty {{{( – 1)}^{k + 1}} \times \frac{k}{{2{k^2} + 1}}} \).

(i) Show that the series is convergent.

(ii) Show that the sum to infinity of the series is less than 0.25.

**Answer/Explanation**

## Markscheme

(i) consider \(\frac{{{T_{n + 1}}}}{{{T_n}}} = \frac{{\left| {\frac{{(n – 1){x^{n + 1}}}}{{{2^{n + 1}}}}} \right|}}{{\left| {\frac{{n{x^n}}}{{{2^n}}}} \right|}}\) *M1*

\( = \frac{{(n + 1)\left| x \right|}}{{2n}}\) *A1*

\( \to \frac{{\left| x \right|}}{2}{\text{ as }}n \to \infty \) *A1*

the radius of convergence satisfies

\(\frac{R}{2} = 1\), *i.e. R* = 2 *A1*

* *

(ii) the series converges for \( – 2 < x < 2\), we need to consider \(x = \pm 2\) *(R1)*

when *x* = 2, the series is \(1 + 2 + 3 + \ldots \) *A1*

this is divergent for any one of several reasons *e.g.* finding an expression for or a comparison test with the harmonic series or noting that \(\mathop {\lim }\limits_{n \to \infty } {u_n} \ne 0\) etc. *R1*

when *x* = – 2, the series is \( – 1 + 2 – 3 + 4 \ldots \) *A1*

this is divergent for any one of several reasons

*e.g.* partial sums are

\( – 1,{\text{ }}1,{\text{ }} – 2,{\text{ }}2,{\text{ }} – 3,{\text{ }}3 \ldots \) or noting that \(\mathop {\lim }\limits_{n \to \infty } {u_n} \ne 0\) *etc.* *R1*

the interval of convergence is \( – 2 < x < 2\) *A1*

*[10 marks]*

(i) this alternating series is convergent because the moduli of successive terms are monotonic decreasing *R1*

and the \({n^{{\text{th}}}}\) term tends to zero as \(n \to \infty \) *R1*

* *

(ii) consider the partial sums

0.333, 0.111, 0.269, 0.148, 0.246 *M1A1*

since the sum to infinity lies between any pair of successive partial sums, it follows that the sum to infinity lies between 0.148 and 0.246 so that it is less than 0.25 *R1*

**Note:** Accept a solution which looks only at 0.333, 0.269, 0.246 and states that these are successive upper bounds.

*[5 marks]*

## Examiners report

Most candidates found the radius of convergence correctly but examining the situation when \(x = \pm 2\) often ended in loss of marks through inadequate explanations. In (b)(i) many candidates were able to justify the convergence of the given series. In (b)(ii), however, many candidates seemed unaware of the fact the sum to infinity lies between any pair of successive partial sums.

Most candidates found the radius of convergence correctly but examining the situation when \(x = \pm 2\) often ended in loss of marks through inadequate explanations. In (b)(i) many candidates were able to justify the convergence of the given series. In (b)(ii), however, many candidates seemed unaware of the fact the sum to infinity lies between any pair of successive partial sums.

## Question

The exponential series is given by \({{\text{e}}^x} = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} \) .

Find the set of values of *x* for which the series is convergent.

(i) Show, by comparison with an appropriate geometric series, that

\[{{\text{e}}^x} – 1 < \frac{{2x}}{{2 – x}},{\text{ for }}0 < x < 2{\text{.}}\]

(ii) Hence show that \({\text{e}} < {\left( {\frac{{2n + 1}}{{2n – 1}}} \right)^n}\), for \(n \in {\mathbb{Z}^ + }\).

(i) Write down the first three terms of the Maclaurin series for \(1 – {{\text{e}}^{ – x}}\) and explain why you are able to state that

\[1 – {{\text{e}}^{ – x}} > x – \frac{{{x^2}}}{2},{\text{ for }}0 < x < 2.\]

(ii) Deduce that \({\text{e}} > {\left( {\frac{{2{n^2}}}{{2{n^2} – 2n + 1}}} \right)^n}\), for \(n \in {\mathbb{Z}^ + }\).

Letting *n* = 1000, use the results in parts (b) and (c) to calculate the value of e correct to as many decimal places as possible.

**Answer/Explanation**

## Markscheme

using a ratio test,

\(\left| {\frac{{{T_{n + 1}}}}{{{T_n}}}} \right| = \left| {\frac{{{x^{n + 1}}}}{{(n + 1)!}}} \right| \times \left| {\frac{{n!}}{{{x^n}}}} \right| = \frac{{\left| x \right|}}{{n + 1}}\) *M1A1*

**Note:** Condone omission of modulus signs.

\( \to 0{\text{ as }}n \to \infty \) for all values of *x* *R1*

the series is therefore convergent for \(x \in \mathbb{R}\) *A1*

*[4 marks]*

(i) \({{\text{e}}^x} – 1 = x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{{2 \times 3}} + …\) *M1*

\( < x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{{2 \times 2}} + …\,\,\,\,\,({\text{for }}x > 0)\) *A1*

\( = \frac{x}{{1 – \frac{x}{2}}}\,\,\,\,\,({\text{for }}x < 2)\) *A1*

\( = \frac{{2x}}{{2 – x}}\,\,\,\,\,({\text{for }}0 < x < 2)\) *AG*

* *

(ii) \({{\text{e}}^x} < 1 + \frac{{2x}}{{2 – x}} = \frac{{2 + x}}{{2 – x}}\) *A1*

\({\text{e}} < {\left( {\frac{{2 + x}}{{2 – x}}} \right)^{\frac{1}{x}}}\) *A1*

replacing *x* by \(\frac{1}{n}\) (and noting that the result is true for \(n > \frac{1}{2}\) and therefore \({\mathbb{Z}^ + }\) ) *M1*

\({\text{e}} < {\left( {\frac{{2n + 1}}{{2n – 1}}} \right)^n}\) *AG*

*[6 marks]*

(i) \(1 – {{\text{e}}^{ – x}} = x – \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} + …\) *A1*

for \(0 < x < 2\), the series is alternating with decreasing terms so that the sum is greater than the sum of an even number of terms *R1*

therefore

\(1 – {{\text{e}}^{ – x}} > x – \frac{{{x^2}}}{2}\) *AG*

* *

(ii) \({{\text{e}}^{ – x}} < 1 – x + \frac{{{x^2}}}{2}\)

\({{\text{e}}^x} > \frac{1}{{\left( {1 – x + \frac{{{x^2}}}{2}} \right)}}\) *M1*

\({\text{e}} > {\left( {\frac{2}{{2 – 2x + {x^2}}}} \right)^{\frac{1}{x}}}\) *A1*

replacing *x* by \(\frac{1}{n}\) (and noting that the result is true for \(n > \frac{1}{2}\) and therefore \({\mathbb{Z}^ + }\) )

\({\text{e}} > {\left( {\frac{{2{n^2}}}{{2{n^2} – 2n + 1}}} \right)^n}\) *AG*

* *

*[4 marks]*

from (b) and (c), \({\text{e}} < 2.718282…\) and \({\text{e}} > 2.718281…\) *A1*

we conclude that e = 2.71828 correct to 5 decimal places *A1*

*[2 marks]*

## Examiners report

Solutions to (a) were generally good although some candidates failed to reach the correct conclusion from correct application of the ratio test. Solutions to (b) and (c), however, were generally disappointing with many candidates unable to make use of the signposting in the question. Candidates who were unable to solve (b) and (c) often picked up marks in (d).

Solutions to (a) were generally good although some candidates failed to reach the correct conclusion from correct application of the ratio test. Solutions to (b) and (c), however, were generally disappointing with many candidates unable to make use of the signposting in the question. Candidates who were unable to solve (b) and (c) often picked up marks in (d).

Solutions to (a) were generally good although some candidates failed to reach the correct conclusion from correct application of the ratio test. Solutions to (b) and (c), however, were generally disappointing with many candidates unable to make use of the signposting in the question. Candidates who were unable to solve (b) and (c) often picked up marks in (d).

## Question

Consider the series \(\sum\limits_{n = 1}^\infty {{{( – 1)}^n}\frac{{{x^n}}}{{n \times {2^n}}}} \).

Find the radius of convergence of the series.

Hence deduce the interval of convergence.

**Answer/Explanation**

## Markscheme

using the ratio test (and absolute convergence implies convergence) *(M1)*

\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{u_{n + 1}}}}{{{u_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\frac{{{{( – 1)}^{n + 1}}{x^{n + 1}}}}{{(n + 1){2^{n + 1}}}}}}{{\frac{{{{( – 1)}^n}{x^n}}}{{(n){2^n}}}}}} \right|\) *A1A1*

**Note:** Award ** A1** for numerator,

**for denominator.**

*A1*

\( = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{( – 1)}^{n + 1}} \times {x^{n + 1}} \times n \times {2^n}}}{{{{( – 1)}^n} \times (n + 1) \times {2^{n + 1}} \times {x^n}}}} \right|\)

\( = \mathop {\lim }\limits_{n \to \infty } \frac{n}{{2(n + 1)}}\left| x \right|\) *(A1)*

\( = \frac{{\left| x \right|}}{2}\) *A1*

for convergence we require \(\frac{{\left| x \right|}}{2} < 1\) *M1*

\( \Rightarrow \left| x \right| < 2\)

hence radius of convergence is 2 *A1*

*[7 marks]*

we now need to consider what happens when \(x = \pm 2\) *(M1)*

when *x* = 2 we have \(\sum\limits_{n = 1}^\infty {\frac{{{{( – 1)}^n}}}{n}} \) which is convergent (by the alternating series test) *A1*

when *x* = −2 we have \(\sum\limits_{n = 1}^\infty {\frac{1}{n}} \) which is divergent *A1*

hence interval of convergence is \(] – 2,{\text{ }}2]\) *A1*

*[4 marks]*

## Examiners report

Most candidates were able to start (a) and a majority gained a fully correct answer. A number of candidates were careless with using the absolute value sign and with dealing with the negative signs and in the more extreme cases this led to candidates being penalised. Part (b) caused more difficulties, with many candidates appearing to know what to do, but then not succeeding in doing it or in not understanding the significance of the answer gained.

Most candidates were able to start (a) and a majority gained a fully correct answer. A number of candidates were careless with using the absolute value sign and with dealing with the negative signs and in the more extreme cases this led to candidates being penalised. Part (b) caused more difficulties, with many candidates appearing to know what to do, but then not succeeding in doing it or in not understanding the significance of the answer gained.

## Question

Consider the infinite series \(\sum\limits_{n = 1}^\infty {\frac{{{n^2}}}{{{2^n}}}{x^n}} \).

Find the radius of convergence.

Find the interval of convergence.

Given that *x* = – 0.1, find the sum of the series correct to three significant figures.

**Answer/Explanation**

## Markscheme

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{u_{n + 1}}}}{{{u_n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{{{{(n + 1)}^2}{x^{n + 1}}}}{{{2^{n + 1}}}}}}{{\frac{{{n^2}{x^n}}}{{{2^n}}}}}\) *M1*

\( = \mathop {\lim }\limits_{n \to \infty } \frac{{{{(n + 1)}^2}}}{{{n^2}}} \times \frac{x}{2}\) *A1*

\( = \frac{x}{2}\) (since \(\lim \to \frac{x}{2}{\text{ as }}n \to \infty \)) *A1*

the radius of convergence *R* is found by equating this limit to 1, giving *R* = 2 *A1*

*[4 marks]*

when *x* = 2, the series is \(\sum {{n^2}} \) which is divergent because the terms do not converge to 0 *R1*

when *x* = –2, the series is \(\sum {{{( – 1)}^n}{n^2}} \) which is divergent because the terms do not converge to 0 *R1*

the interval of convergence is \(] – 2,{\text{ }}2[\) *A1*

*[3 marks]*

putting *x* = – 0.1, *(M1)*

for any correct partial sum *(A1)*

– 0.05

– 0.04

– 0.041125

– 0.041025

– 0.0410328 *(A1)*

the sum is – 0.0410 correct to 3 significant figures *A1*

*[4 marks]*

## Examiners report

It was pleasing that most candidates were aware of the Radius of Convergence and Interval of Convergence required by parts (a) and (b) of this problem. Many candidates correctly handled the use of the Ratio Test for convergence and there was also the use of Cauchy’s n^{th} root test by a small number of candidates to solve part (a). Candidates need to take care to justify correctly the divergence or convergence of series when finding the Interval of Convergence.

It was pleasing that most candidates were aware of the Radius of Convergence and Interval of Convergence required by parts (a) and (b) of this problem. Many candidates correctly handled the use of the Ratio Test for convergence and there was also the use of Cauchy’s n^{th} root test by a small number of candidates to solve part (a). Candidates need to take care to justify correctly the divergence or convergence of series when finding the Interval of Convergence.

The summation of the series in part (c) was poorly handled by a significant number of candidates, which was surprising on what was expected to be quite a straightforward problem. Again efficient use of the GDC seemed to be a problem. A number of candidates found the correct sum but not to the required accuracy.

Marks available | 9 |

Reference code | 13N.3ca.hl.TZ0.5 |

## Question

A function \(f\) is defined in the interval \(\left] { – k,{\text{ }}k} \right[\), where \(k > 0\). The gradient function \({f’}\) exists at each point of the domain of \(f\).

The following diagram shows the graph of \(y = f(x)\), its asymptotes and its vertical symmetry axis.

(a) Sketch the graph of \(y = f'(x)\).

Let \(p(x) = a + bx + c{x^2} + d{x^3} + \ldots \) be the Maclaurin expansion of \(f(x)\).

(b) (i) Justify that \(a > 0\).

(ii) Write down a condition for the largest set of possible values for each of the parameters \(b\), \(c\) and \(d\).

(c) State, with a reason, an upper bound for the radius of convergence.

**Answer/Explanation**

## Markscheme

(a)

** A1** for shape,

**for passing through origin**

*A1*

*A1A1***Note: **Asymptotes not required.

*[2 marks]*

(b) \(p(x) = \underbrace {f(0)}_a + \underbrace {f'(0)}_bx + \underbrace {\frac{{f”(0)}}{{2!}}}_c{x^2} + \underbrace {\frac{{{f^{(3)}}(0)}}{{3!}}}_d{x^3} + \ldots \)

(i) because the *y*-intercept of \(f\) is positive *R1*

(ii) \(b = 0\) *A1*

\(c \geqslant 0\) *A1A1*

**Note: A1 **for \( > \) and

**for \( = \).**

*A1*\(d = 0\) *A1*

*[5 marks]*

(c) as the graph has vertical asymptotes \(x = \pm k,{\text{ }}k > 0\), *R1*

the radius of convergence has an upper bound of \(k\) *A1*

**Note: **Accept \(r < k\).

*[2 marks]*

## Examiners report

Overall candidates made good attempts to parts (a) and most candidates realized that the graph contained the origin; however many candidates had difficulty rendering the correct shape of the graph of \(f’\). Part b(i) was also well answered although some candidates where not very clear and digressed a lot. Part (ii) was less successful with most candidates scoring just part of the marks. A small number of candidates answered part (c) correctly with a valid reason.

## Question

Each term of the power series \(\frac{1}{{1 \times 2}} + \frac{1}{{4 \times 5}}x + \frac{1}{{7 \times 8}}{x^2} + \frac{1}{{10 \times 11}}{x^3} + \ldots \) has the form \(\frac{1}{{b(n) \times c(n)}}{x^n}\), where \(b(n)\) and \(c(n)\) are linear functions of \(n\).

(a) Find the functions \(b(n)\) and \(c(n)\).

(b) Find the radius of convergence.

(c) Find the interval of convergence.

**Answer/Explanation**

## Markscheme

(a) \(b(n) = 3n + 1\) *A1*

\(c(n) = 3n + 2\) *A1*

**Note: **\(b(n)\) and \(c(n)\) may be reversed.

*[2 marks]*

(b) consider the ratio of the \({(n + 1)^{{\text{th}}}}\) and \({n^{{\text{th}}}}\) terms: *M1*

\(\frac{{3n + 1}}{{3n + 4}} \times \frac{{3n + 2}}{{3n + 5}} \times \frac{{{x^{n + 1}}}}{{{x^n}}}\) **A1**

\(\mathop {{\text{lim}}}\limits_{n \to 0} \frac{{3n + 1}}{{3n + 4}} \times \frac{{3n + 2}}{{3n + 5}} \times \frac{{{x^{n + 1}}}}{{{x^n}}}x\) *A1*

radius of convergence: \(R = 1\) *A1*

*[4 marks]*

(c) any attempt to study the series for \(x = -1\) or \(x = 1\) *(M1)*

converges for \(x = 1\) by comparing with *p*-series \(\sum {\frac{1}{{{n^2}}}} \) *R1*

attempt to use the alternating series test for \(x = -1\) *(M1)*

**Note: **At least one of the conditions below needs to be attempted for *M1*.

\(\left| {{\text{terms}}} \right| \approx \frac{1}{{9{n^2}}} \to 0\) and terms decrease monotonically in absolute value **A1**

series converges for \(x = -1\) *R1*

interval of convergence: \(\left[ { – 1,{\text{ 1}}} \right]\) *A1*

**Note: **Award the ** R1**s only if an attempt to corresponding correct test is made;

award the final ** A1 **only if at least one of the

**s is awarded;**

*R1*Accept study of absolute convergence at end points.

*[6 marks]*

## Examiners report

## Question

Consider the function \(f(x) = \frac{1}{{1 + {x^2}}},{\text{ }}x \in \mathbb{R}\).

Illustrate graphically the inequality, \(\frac{1}{5}\sum\limits_{r = 1}^5 {f\left( {\frac{r}{5}} \right) < \int_0^1 {f(x){\text{d}}x < \frac{1}{5}\sum\limits_{r = 0}^4 {f\left( {\frac{r}{5}} \right)} } } \).

Use the inequality in part (a) to find a lower and upper bound for \(\pi \).

Show that \(\sum\limits_{r = 0}^{n – 1} {{{( – 1)}^r}{x^{2r}} = \frac{{1 + {{( – 1)}^{n – 1}}{x^{2n}}}}{{1 + {x^2}}}} \).

Hence show that \(\pi = 4\left( {\sum\limits_{r = 0}^{n – 1} {\frac{{{{( – 1)}^r}}}{{2r + 1}} – {{( – 1)}^{n – 1}}\int_0^1 {\frac{{{x^{2n}}}}{{1 + {x^2}}}{\text{d}}x} } } \right)\).

**Answer/Explanation**

## Markscheme

* A1A1A1*

** A1 **for upper rectangles,

**for lower rectangles,**

*A1***for curve in between with \(0 \le x \le 1\)**

*A1*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

Consider the infinite series \(\sum\limits_{n = 1}^\infty {\frac{{(n – 1){x^n}}}{{{n^2} \times {2^n}}}} \) .

Find the radius of convergence.

Find the interval of convergence.

**Answer/Explanation**

## Markscheme

using the ratio test, \(\frac{{{u_{n + 1}}}}{{{u_n}}} = \frac{{n{x^{n + 1}}}}{{{{(n + 1)}^2}{2^{n + 1}}}} \times \frac{{{n^2}{2^n}}}{{(n – 1){x^n}}}\) *M1*

\( = \frac{{{n^3}}}{{{{(n + 1)}^2}(n – 1)}} \times \frac{x}{2}\) *A1*

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{u_{n + 1}}}}{{{u_n}}} = \frac{x}{2}\) *A1*

the radius of convergence *R* satisfies

\(\frac{R}{2} = 1\) so *R* = 2 *A1*

*[4 marks]*

considering *x* = 2 for which the series is

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

using the limit comparison test with the harmonic series *M1*

\(\sum\limits_{n = 1}^\infty {\frac{1}{n}} \), which diverges

consider

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{u_n}}}{{\frac{1}{n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{n – 1}}{n} = 1\) *A1*

the series is therefore divergent for *x* = 2 *A1*

when *x* = –2 , the series is

\(\sum\limits_{n = 1}^\infty {\frac{{(n – 1)}}{{{n^2}}} \times {{( – 1)}^n}} \)

this is an alternating series in which the \({n^{{\text{th}}}}\) term tends to 0 as \(n \to \infty \) *A1*

consider \(f(x) = \frac{{x – 1}}{{{x^2}}}\) *M1*

\(f'(x) = \frac{{2 – x}}{{{x^3}}}\) *A1*

this is negative for \(x > 2\) so the sequence \(\{ |{u_n}|\} \) is eventually decreasing *R1*

the series therefore converges when *x* = –2 by the alternating series test *R1*

the interval of convergence is therefore [–2, 2[ *A1*

*[9 marks]*

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