Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

(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\(\frac{{577}}{{408}}\).

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TABLE OF CONTENTS :

TABLE OF CONTENTS

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, \(\sqrt 2 = 1.41421356\))

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Essential Calculus: Early Transcendentals

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Short Answer

Step by Step Solution

Subpart (a) Given data:-

ApplyingDefinition of limit of a sequence:-

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

Simplifying further for \(n > N\):-

Subpart (b):Given data:-

Substituting \(n = 1\) and \(n = 2\)in the given term:-

Substituting \(n = 3\)and \(n = 4\):-

Substituting \(n = 5\), \(n = 6\) and \(n = 7\):-

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

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Step by Step Solution

Subpart (a) Given data:-

ApplyingDefinition of limit of a sequence:-

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

Simplifying further for \(n > N\):-

Subpart (b):Given data:-

Substituting \(n = 1\) and \(n = 2\)in the given term:-

Substituting \(n = 3\)and \(n = 4\):-

Substituting \(n = 5\), \(n = 6\) and \(n = 7\):-

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

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