The function $f$ is defined by $f(x) = \frac{{{{\rm{e}}^x} + {{\rm{e}}^{ - x}}}}{2}$ .

  (i)     Obtain an expression for ${f^{(n)}}(x)$ , the nth derivative of $f(x)$ with respect to $x$.

  (ii)     Hence derive the Maclaurin series for $f(x)$ up to and including the term in ${x^4}$ .

  (iii)     Use your result to find a rational approximation to $f\left( {\frac{1}{2}} \right)$ .

  (iv)     Use the Lagrange error term to determine an upper bound to the error in this approximation.

Use the integral test to determine whether the series $\sum\limits_{n = 1}^\infty  {\frac{{\ln n}}{{{n^2}}}}$ is convergent or divergent.

(i)     ${f^{(n)}}(x) = \frac{{{{\rm{e}}^x} + {{( - 1)}^n}{{\rm{e}}^{ - x}}}}{2}$     (M1)A1

(ii)     Coefficient of ${x^n} = \frac{{{f^{(n)}}(0)}}{{n!}}$     (M1)

$= \frac{{1 + {{( - 1)}^n}}}{{2n!}}$     (A1)

$f(x) = 1 + \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{24}} +  \ldots$     A1

(iii)     Putting $x = \frac{1}{2}$     M1

$f(0.5) = 1 + \frac{1}{8} + \frac{1}{{16 \times 24}} = \frac{{433}}{{384}}$     (M1)A1

(iv)     Lagrange error term $= \frac{{{f^{(n + 1)}}(c)}}{{(n + 1)!}}{x^{n + 1}}$     M1

$= \frac{{{f^{(5)}}(c)}}{{120}} \times {\left( {\frac{1}{2}} \right)^5}$     A1

${{f^{(5)}}(c)}$ is an increasing function because – any valid reason, e.g. plotted a graph, positive derivative, increasing function minus a decreasing function, so this is maximized when $x = 0.5$ .     R1

Therefore upper bound $= \frac{{({{\rm{e}}^{0.5}} - {{\rm{e}}^{ - 0.5}})}}{{2 \times 120}} \times {\left( {\frac{1}{2}} \right)^5}$     M1

$= 0.000136$     A1

[13 marks]

We consider $\int_1^\infty  {\frac{{\ln x}}{{{x^2}}}} {\rm{d}}x = \int_1^\infty  {\ln x{\rm{d}}x} \left( { - \frac{1}{2}} \right)$     M1A1

$= \left[ { - \frac{{\ln x}}{x}} \right]_1^\infty  + \int_1^\infty  {\frac{{1x}}{{{x^2}}}} {\rm{d}}x$     A1A1

$= \left[ { - \frac{{\ln x}}{x}} \right]_1^\infty  - \left[ {\frac{1}{x}} \right]_1^\infty$     A1

Now $\mathop {\lim }\limits_{x \to \infty } \left( {\frac{1}{x}} \right) = 0$     R1

$\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{\ln x}}{x}} \right) = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{1}{x}} \right) = 0$     M1A1

The integral is convergent with value $1$ and so therefore is the series.      R1

[9 marks]