## 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

Find the set of values of *k *for which the improper integral \(\int_2^\infty {\frac{{{\text{d}}x}}{{x{{(\ln x)}^k}}}} \) converges.

Show that the series \(\sum\limits_{r = 2}^\infty {\frac{{{{( – 1)}^r}}}{{r\ln r}}} \) is convergent but not absolutely convergent.

**Answer/Explanation**

## Markscheme

consider the limit as \(R \to \infty \) of the (proper) integral

\(\int_2^R {\frac{{{\text{d}}x}}{{x{{(\ln x)}^k}}}} \) ** (M1)**

substitute \(u = \ln x,{\text{ d}}u = \frac{1}{x}{\text{d}}x\) (*M1)*

obtain \(\int_{\ln 2}^{\ln R} {\frac{1}{{{u^k}}}{\text{d}}u = \left[ { – \frac{1}{{k – 1}}\frac{1}{{{u^{k – 1}}}}} \right]_{\ln 2}^{\ln R}} \) *A1*** **

**Note: **Ignore incorrect limits or omission of limits at this stage.

or \([\ln u]_{\ln 2}^{\ln R}\) if *k* = 1 *A1*** **

**Note: **Ignore incorrect limits or omission of limits at this stage.

because \(\ln R{\text{ }}({\text{and }}\ln \ln R) \to \infty {\text{ as }}R \to \infty \) *(M1)*

converges in the limit if *k* > 1 *A1*

*[6 marks]*

C: \({\text{terms}} \to 0{\text{ as }}r \to \infty \) *A1*

\(\left| {{u_{r + 1}}} \right| < \left| {{u_r}} \right|\) for all *r **A1*

convergence by alternating series test *R1*

AC: \({(x\ln x)^{ – 1}}\) is positive and decreasing on \([2,\,\infty )\) *A1*

not absolutely convergent by integral test using part (a) for *k* = 1 *R1*

*[5 marks]*

## Examiners report

A good number of candidates were able to find the integral in part (a) although the vast majority did not consider separately the integral when *k* = 1. Many candidates did not explicitly set a limit for the integral to let this limit go to infinity in the anti – derivative and it seemed that some candidates were “substituting for infinity”. This did not always prevent candidates finding a correct final answer but the lack of good technique is a concern. In part (b) many candidates seemed to have some knowledge of the relevant test for convergence but this test was not always rigorously applied. In showing that the series was not absolutely convergent candidates were often not clear in showing that the function being tested had to meet a number of criteria and in so doing lost marks.

A good number of candidates were able to find the integral in part (a) although the vast majority did not consider separately the integral when *k* = 1. Many candidates did not explicitly set a limit for the integral to let this limit go to infinity in the anti – derivative and it seemed that some candidates were “substituting for infinity”. This did not always prevent candidates finding a correct final answer but the lack of good technique is a concern. In part (b) many candidates seemed to have some knowledge of the relevant test for convergence but this test was not always rigorously applied. In showing that the series was not absolutely convergent candidates were often not clear in showing that the function being tested had to meet a number of criteria and in so doing lost marks.

## 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.