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]