By successive differentiation find the first four non-zero terms in the Maclaurin series for \(f(x) = (x + 1)\ln (1 + x) - x\).

Deduce that, for \(n \geqslant 2\), the coefficient of \({x^n}\) in this series is \({( - 1)^n}\frac{1}{{n(n - 1)}}\).

By applying the ratio test, find the radius of convergence for this Maclaurin series.

\(f(x) = (x + 1)\ln (1 + x) - x\)     \(f(0) = 0\)    A1

\(f'(x) = \ln (1 + x) + \frac{{x + 1}}{{1 + x}} - 1{\text{ }}\left( { = \ln (1 + x)} \right)\)     \(f'(0) = 0\)    M1A1A1

\(f''(x) = {(1 + x)^{ - 1}}\)     \(f''(0) = 1\)    A1A1

\(f'''(x) =  - {(1 + x)^{ - 2}}\)     \(f'''(0) =  - 1\)    A1

\({f^{(4)}}(x) = 2{(1 + x)^{ - 3}}\)     \({f^{(4)}}(0) = 2\)    A1

\({f^{(5)}}(x) =  - 3 \times 2{(1 + x)^{ - 4}}\)     \({f^{(5)}}(0) =  - 3 \times 2\)    A1

\(f(x) = \frac{{{x^2}}}{{2!}} - \frac{{1{x^3}}}{{3!}} + \frac{{2{x^4}}}{{4!}} - \frac{{6{x^5}}}{{5!}} \ldots \)    M1A1

\(f(x) = \frac{{{x^2}}}{{1 \times 2}} - \frac{{{x^3}}}{{2 \times 3}} + \frac{{{x^4}}}{{3 \times 4}} - \frac{{{x^5}}}{{4 \times 5}} \ldots \)

\(f(x) = \frac{{{x^2}}}{2} - \frac{{{x^3}}}{6} + \frac{{{x^4}}}{{12}} - \frac{{{x^5}}}{{20}} \ldots \)

Note: Allow follow through from the first error in a derivative (provided future derivatives also include the chain rule), no follow through after a second error in a derivative.

[11 marks]

\({f^{(n)}}(0) = {( - 1)^n}(n - 2)!\) So coefficient of \({x^n} = {( - 1)^n}\frac{{(n - 2)!}}{{n!}}\)     A1

coefficient of \({x^n}\) is \({( - 1)^n}\frac{1}{{n(n - 1)}}\)     AG

[1 mark]

applying the ratio test to the series of absolute terms

\(\mathop {\lim }\limits_{n \to \infty } \frac{{\frac{{{{\left| x \right|}^{n + 1}}}}{{(n + 1)n}}}}{{\frac{{{{\left| x \right|}^n}}}{{n(n - 1)}}}}\)    M1A1

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

\( = \left| x \right|\)    A1

so for convergence \(\left| x \right| < 1\), giving radius of convergence as 1     (M1)A1

[6 marks]