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Found in: Page 435

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# (a)Show that if $$\mathop {\lim }\limits_{n \to \infty } {a_2}_n = L$$and $$\mathop {\lim }\limits_{n \to \infty } {a_{2n + 1}} = L,$$ then {$${a_n}$$} is convergent and $$\mathop {\lim }\limits_{n \to \infty } {a_n} = L$$.(a) If $${a_1} = 1$$ and $${a_{n + 1}} = 1 + \frac{1}{{1 + {a_n}}}$$Find the first eight terms of the sequence {$${a_n}$$}. Then use part(a) to show that $$\mathop {\lim }\limits_{n \to \infty } {a_n} = \sqrt 2$$. This gives the continued fraction expansion $$\sqrt 2 = 1 + \frac{1}{{2 + \frac{1}{{2 + ...}}}}$$

(a) Using limit of a sequence definition, it is shown that $$\mathop {\lim }\limits_{n \to \infty } {a_n} = L$$ and {$${a_n}$$} is convergent.

(b) The first eight terms of the sequence {$${a_n}$$} are, $$1,\frac{3}{2},\frac{7}{5},\frac{{17}}{{12}},\frac{{41}}{{29}},\frac{{99}}{{70}},\frac{{239}}{{169}}$$and$$\frac{{577}}{{408}}$$.

See the step by step solution

## Subpart (a) Given data:-

The sub sequences {$${a_{2n}}$$} and {$${a_{2n + 1}}$$} are converges to $$L$$. i.e., $$\mathop {\lim }\limits_{n \to \infty } {a_{2n}} = L$$and $$\mathop {\lim }\limits_{n \to \infty } {a_{2n + 1}} = L.$$

## ApplyingDefinition of limit of a sequence:-

We have,

$$\Rightarrow \mathop {\lim }\limits_{n \to \infty } {a_n}$$

from the limit of a sequence definition, for every $$\varepsilon > 0,$$there exists a positive integer $$N$$such that for $$n > N,$$

$$\left| {{a_n} - L} \right| < \varepsilon$$.

## Applying limit of a sequence to the given sub sequences {$${a_{2n}}$$} and {$${a_{2n + 1}}$$ } :-

As $$\mathop {\lim }\limits_{n \to \infty } {a_{2n}} = L$$and $$\mathop {\lim }\limits_{n \to \infty } {a_{2n + 1}} = L$$, then for every $$\varepsilon > 0$$, there exists a positive integer

$${N_1}$$, such that for $$n > {N_1},$$

$$\left| {{a_{2n}} - L} \right| < \varepsilon$$.

Also, for every $$\varepsilon > 0,$$there exists a positive integer $${N_2}$$ such that for $$n > {N_2},$$

$$\left| {{a_{2n + 1}} - L} \right| < \varepsilon$$.

## Simplifying further for $$n > N$$:-

Consider $$N = \max \{ {N_1},2{N_1} + 1\}$$

For $$n > N,$$

$$n = 2k$$or $$n = 2k + 1$$, for $$k \in N.$$

If $$n = 2k$$, then $$\left| {{a_{2k}} - L} \right| < \varepsilon$$as $$k > {N_1}.$$

If $$n = 2k + 1$$, then $$\left| {{a_{2k + 1}} - L} \right| < \varepsilon$$as $$k > {N_2}.$$

Therefore, the sequence {$${a_n}$$} is convergent, and $$\mathop {\lim }\limits_{n \to \infty } {a_n} = L.$$

## Subpart (b):Given data:-

First term of the sequence,

$${a_1} = 1,$$and,

$${a_{n + 1}} = 1 + \frac{1}{{1 + {a_n}}}$$.

## Substituting $$n = 1$$ and $$n = 2$$in the given term:-

For $$n = 1$$ to the term $${a_{n + 1}} = 1 + \frac{1}{{1 + {a_n}}}$$,

$${a_{1 + 1}} = 1 + \frac{1}{{1 + {a_1}}}$$

$${a_2} = 1 + \frac{1}{{1 + 1}} = 1 + \frac{1}{2} = \frac{3}{2}$$

For $$n = 2,$$

\begin{aligned}{a_{2 + 1}} = 1 + &\frac{1}{{1 + {a_2}}}\\{a_3} = 1 + &\frac{1}{{1 + \frac{3}{2}}}\\{a_3} = 1 + &\frac{2}{5}\\{a_3} =&\frac{7}{5}.\end{aligned}

## Substituting $$n = 3$$and $$n = 4$$:-

For $$n = 3,$$

\begin{aligned}{}{a_{3 + 1}} = 1 + \frac{1}{{1 + {a_3}}}\\{a_4} = 1 + \frac{1}{{1 + \frac{7}{5}}}\\{a_4} = 1 + \frac{5}{{12}}\\{a_4} = \frac{{17}}{{12}}.\end{aligned}

For $$n = 4,$$

\begin{aligned}{}{a_{4 + 1}} = 1 + \frac{1}{{1 + {a_4}}}\\{a_5} = 1 + \frac{1}{{1 + \frac{{17}}{{12}}}}\\{a_5} = 1 + \frac{{12}}{{29}}\\{a_5} = \frac{{41}}{{29}}.\end{aligned}

## Substituting $$n = 5$$, $$n = 6$$ and $$n = 7$$:-

For $$n = 5,$$

\begin{aligned}{}{a_{5 + 1}} = 1 + \frac{1}{{1 + {a_5}}}\\{a_6} = 1 + \frac{1}{{1 + \frac{{41}}{{29}}}}\\{a_6} = 1 + \frac{{29}}{{70}}\\{a_6} = \frac{{99}}{{70}}.\end{aligned}

For $$n = 6,$$

\begin{aligned}{}{a_{6 + 1}} = 1 + \frac{1}{{1 + {a_6}}}\\{a_7} = 1 + \frac{1}{{1 + \frac{{99}}{{70}}}}\\{a_7} = 1 + \frac{{70}}{{169}}\\{a_7} = \frac{{239}}{{169}}.\end{aligned}

For $$n = 7,$$

\begin{aligned}{}{a_{7 + 1}} = 1 + \frac{1}{{1 + {a_7}}}\\{a_8} = 1 + \frac{1}{{1 + \frac{{239}}{{169}}}}\\{a_8} = 1 + \frac{{169}}{{408}}\\{a_8} = \frac{{577}}{{408}}.\end{aligned}

Therefore, the first eight terms of the sequence are$$1,\frac{3}{2},\frac{7}{5},\frac{{17}}{{12}},\frac{{41}}{{29}},\frac{{99}}{{70}},\frac{{239}}{{169}},\frac{{577}}{{408}}$$

## Using part(a) for showing $$\mathop {\lim }\limits_{n \to \infty } {a_n} = \sqrt 2$$

Consider the two sub sequences of {$${a_n}$$} as,

Odd terms – $$\{ 1,\frac{7}{5},\frac{{41}}{{29}},\frac{{239}}{{169}},...\}$$and

Even terms - $$\{ \frac{3}{2},\frac{{17}}{{12}},\frac{{99}}{{70}},\frac{{577}}{{408}},...\}$$

Odd terms sequence is increasing and tending to $$1.41421356$$.

Even terms sequence is decreasing and tending to $$1.41421356$$.

Therefore, the sequence, $${a_n}$$approaches to $$\sqrt 2$$ (Since, $$\sqrt 2 = 1.41421356$$)