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.

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]

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).

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).