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The general term of a sequence $\left\{a_{n}\right\}$ is given by the formula $a_{n}=\frac{\mathrm{e}^{n}+2^{n}}{2 \mathrm{e}^{n}}, n \in \mathbb{Z}^{+}$ .

(a)

Show that the sequence $\left\{a_{n}\right\}$ is convergent and find the limit L.

[ 2 ]

$\quad L=\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} \frac{1}{2}+\frac{1}{2}\left(\frac{2}{\mathrm{e}}\right)^{n}=\frac{1}{2}+\frac{1}{2} \times 0=\frac{1}{2}$

(b)

Find the smallest value of $N \in \mathbb{Z}^{+}$ such that $\left|a_{n}-L\right|<0.001$ , for all $n \geq N$ .

[ 4 ]

$\quad\left|a_{n}-\frac{1}{2}\right|=\left|\frac{1}{2}+\frac{1}{2}\left(\frac{2}{\mathrm{e}}\right)^{n}-\frac{1}{2}\right|=\left|\frac{1}{2}\left(\frac{2}{\mathrm{e}}\right)^{n}\right|<\frac{1}{1000}$

EITHER
$\Rightarrow\left(\frac{\mathrm{e}}{2}\right)^{n}>500$ $\Rightarrow n>20.25 \ldots$

OR
$\Rightarrow\left(\frac{2}{\mathrm{e}}\right)^{n}<500$ $\Rightarrow n>20.25 \ldots$

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

THEN
therefore N=21

IB Maths AA HL Ahl 5 12 Hl Continuity Differentiability and First Principles Topic Practice

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IB Maths AA HL Ahl 5 12 Hl Continuity Differentiability and First Principles Topic Practice

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Question 1

Question (a)

Question (b)