The general term of a sequence \(\{ {a_n}\} \) is given by the formula \({a_n} = \frac{{{{\text{e}}^n} + {2^n}}}{{2{{\text{e}}^n}}},{\text{ }}n \in {\mathbb{Z}^ + }\).

(a)     Determine whether the sequence \(\{ {a_n}\} \) is decreasing or increasing.

(b)     Show that the sequence \(\{ {a_n}\} \) is convergent and find the limit L.

(c)     Find the smallest value of \(N \in {\mathbb{Z}^ + }\) such that \(\left| {{a_n} - L} \right| < 0.001\), for all \(n \geqslant N\).

(a)     \({a_n} = \frac{{{{\text{e}}^n} + {2^n}}}{{2{{\text{e}}^n}}} = \frac{1}{2} + \frac{1}{2}{\left( {\frac{2}{{\text{e}}}} \right)^2} > \frac{1}{2} + \frac{1}{2}{\left( {\frac{2}{{\text{e}}}} \right)^{n + 1}} = {a_{n + 1}}\)     M1A1

the sequence is decreasing (as terms are positive)     A1

Note: Accept reference to the sum of a constant and a decreasing geometric sequence.

Note: Accept use of derivative of \(f(x) = \frac{{{{\text{e}}^x} + 2x}}{{2{{\text{e}}^x}}}\) (and condone use of n) and graphical methods (graph of the sequence or graph of corresponding function \(f\) or graph of its derivative \({f'}\)).

     Accept a list of consecutive terms of the sequence clearly decreasing (eg \(0.8678 \ldots ,{\text{ }}0.77067 \ldots ,{\text{ }} \ldots \)).

[3 marks]

(b)     \(L = \mathop {\lim }\limits_{n \to \infty } {a_n} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{2} + \frac{1}{2}{\left( {\frac{2}{{\text{e}}}} \right)^n} = \frac{1}{2} + \frac{1}{2} \times 0 = \frac{1}{2}\)     M1A1

[2 marks]

(c)     \(\left| {{a_n} - \frac{1}{2}} \right| = \left| {\frac{1}{2} + \frac{1}{2}{{\left( {\frac{2}{{\text{e}}}} \right)}^n} - \frac{1}{2}} \right| = \left| {\frac{1}{2}{{\left( {\frac{2}{{\text{e}}}} \right)}^n}} \right| < \frac{1}{{1000}}\)     M1

EITHER

\( \Rightarrow {\left( {\frac{{\text{e}}}{2}} \right)^n} > 500\)     (A1)

\( \Rightarrow n > 20.25 \ldots \)     (A1)

OR

\( \Rightarrow {\left( {\frac{2}{{\text{e}}}} \right)^n} < 500\)

\( \Rightarrow n > 20.25 \ldots \)     (A1)(A1)

Note: A1 for correct inequality; A1 for correct value.

THEN

therefore \(N = 21\)     A1

[4 marks]

Most candidates were successful in answering part (a) using a variety of methods. The majority of candidates scored some marks, if not full marks. Surprisingly, some candidates did not have the correct graph for the function the sequence represents. They obviously did not enter it correctly into their GDCs. Others used one of the two definitions for showing that a sequence is increasing/decreasing, but made mistakes with the algebraic manipulation of the expression, thereby arriving at an incorrect answer. Part (b) was less well answered with many candidates ignoring the command terms ‘show that’ and ‘find’ and just writing down the value of the limit. Some candidates attempted to use convergence tests for series with this sequence. Part (c) of this question was found challenging by the majority of candidates due to difficulties in solving inequalities involving absolute value.