# IB DP Maths Topic 9.1 Infinite sequences of real numbers and their convergence or divergence. HL Paper 3

## Question

The sequence $$\{ {u_n}\}$$ is defined by $${u_n} = \frac{{3n + 2}}{{2n – 1}}$$, for $$n \in {\mathbb{Z}^ + }$$.

Show that the sequence converges to a limit L , the value of which should be stated.

[3]
a.

Find the least value of the integer N such that $$\left| {{u_n} – L} \right| < \varepsilon$$ , for all n > N where

(i)     $$\varepsilon = 0.1$$;

(ii)     $$\varepsilon = 0.00001$$.

[4]
b.

For each of the sequences $$\left\{ {\frac{{{u_n}}}{n}} \right\},{\text{ }}\left\{ {\frac{1}{{2{u_n} – 2}}} \right\}$$ and $$\left\{ {{{( – 1)}^n}{u_n}} \right\}$$ , determine whether or not it converges.

[6]
c.

Prove that the series $$\sum\limits_{n = 1}^\infty {({u_n} – L)}$$ diverges.

[2]
d.

## Markscheme

$${u_n} = \frac{{3 + \frac{2}{n}}}{{2 – \frac{1}{n}}}$$ or $$\frac{3}{2} + \frac{A}{{2n – 1}}$$     M1

using $$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n} = 0$$     (M1)

obtain $$\mathop {\lim }\limits_{n \to \infty } {u_n} = \frac{3}{2} = L$$     A1     N1

[3 marks]

a.

$${u_n} – L = \frac{7}{{2(2n – 1)}}$$     (A1)

$$\left| {{u_n} – L} \right| < \varepsilon \Rightarrow n > \frac{1}{2}\left( {1 + \frac{7}{{2\varepsilon }}} \right)$$     (M1)

(i)     $$\varepsilon = 0.1 \Rightarrow N = 18$$     A1

(ii)     $$\varepsilon = 0.00001 \Rightarrow N = 175000$$     A1

[4 marks]

b.

$${u_n} \to L$$ and $$\frac{1}{n} \to 0$$     M1

$$\Rightarrow \frac{{{u_n}}}{n} \to (L \times 0) = 0$$ , hence converges     A1

$$2{u_n} – 2 \to 2L – 2 = 1 \Rightarrow \frac{1}{{2{u_n} – 2}} \to 1$$ , hence converges     M1A1

Note: To award A1 the value of the limit and a statement of convergence must be clearly seen for each sequence.

$${( – 1)^n}{u_n}$$ does not converge     A1

The sequence alternates (or equivalent wording) between values close to $$\pm L$$     R1

[6 marks]

c.

$${u_n} – L > \frac{7}{{4n}}$$ (re: harmonic sequence)     M1

$$\Rightarrow \sum\limits_{n = 1}^\infty {({u_n} – L)}$$ diverges by the comparison theorem     R1

Note: Accept alternative methods.

[2 marks]

d.

## Examiners report

The “show that” in part (a) of this problem was not adequately dealt with by a significant minority of candidates and simply stating the limit and not demonstrating its existence lost marks. Part (b), whilst being possible without significant knowledge of limits, seemed to intimidate some candidates due to its unfamiliarity and the notation. Part (c) was somewhat disappointing as many candidates attempted to apply rules on the convergence of series to solve a problem that was dealing with the limits of sequences. The same confusion was seen on part (d) where also some errors in algebra prevented candidates from achieving full marks.

a.

The “show that” in part (a) of this problem was not adequately dealt with by a significant minority of candidates and simply stating the limit and not demonstrating its existence lost marks. Part (b), whilst being possible without significant knowledge of limits, seemed to intimidate some candidates due to its unfamiliarity and the notation. Part (c) was somewhat disappointing as many candidates attempted to apply rules on the convergence of series to solve a problem that was dealing with the limits of sequences. The same confusion was seen on part (d) where also some errors in algebra prevented candidates from achieving full marks.

b.

The “show that” in part (a) of this problem was not adequately dealt with by a significant minority of candidates and simply stating the limit and not demonstrating its existence lost marks. Part (b), whilst being possible without significant knowledge of limits, seemed to intimidate some candidates due to its unfamiliarity and the notation. Part (c) was somewhat disappointing as many candidates attempted to apply rules on the convergence of series to solve a problem that was dealing with the limits of sequences. The same confusion was seen on part (d) where also some errors in algebra prevented candidates from achieving full marks.

c.

The “show that” in part (a) of this problem was not adequately dealt with by a significant minority of candidates and simply stating the limit and not demonstrating its existence lost marks. Part (b), whilst being possible without significant knowledge of limits, seemed to intimidate some candidates due to its unfamiliarity and the notation. Part (c) was somewhat disappointing as many candidates attempted to apply rules on the convergence of series to solve a problem that was dealing with the limits of sequences. The same confusion was seen on part (d) where also some errors in algebra prevented candidates from achieving full marks.

d.

## Question

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 $$n \geqslant N$$.

## Markscheme

(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]

## Examiners report

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.