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Q17E

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Essential Calculus: Early Transcendentals
Found in: Page 434
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

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Short Answer

Determine whether the sequence converges or diverges. If it converges, find the limit.

\({a_n} = \frac{{{{( - 1)}^n}}}{{2\sqrt n }}\)

The sequence converges and the value of limit is 0.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Definition

A sequence\(\left\{ {{a_n}} \right\}\)has the limit \(L\)and we write \(\mathop {\lim }\limits_{n \to \infty } {a_n} = L\;\;or\;\;{a_n} \to L\)as\(n \to \infty \)if we can make the terms\({a_n}\)as close to\(L\)as we like by taking\(n\)sufficiently large. If \(\mathop {\lim }\limits_{n \to \infty } {a_n}\)exists, we say the sequence converges (or is convergent). Otherwise, we say the sequence diverges (or is divergent).

Simplify

Given a sequence \({a_n} = \frac{{{{( - 1)}^n}}}{{2\sqrt n }}\).

Evaluate\(\mathop {\lim }\limits_{n \to \infty } |{a_n}|\) as:

\(\begin{aligned}\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{( - 1)}^n}}}{{2\sqrt n }}} \right| &= \frac{1}{2}\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{{\sqrt n }}} \right)\\ &= \frac{1}{2} \cdot 0\\ &= 0\end{aligned}\)

Compute\(\mathop {\lim }\limits_{n \to \infty } {a_n}\)

Now using the theorem:

If\(\mathop {\lim }\limits_{n \to \infty } \left| {{a_n}} \right| = 0\), then\(\mathop {\lim }\limits_{n \to \infty } {a_n} = 0\)

\(\begin{aligned}\mathop {\lim }\limits_{n \to \infty } {a_n} &= \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{( - 1)}^n}}}{{2\sqrt n }}} \right)\\ &= 0\end{aligned}\).

So, the sequence \({a_n} = \frac{{{{( - 1)}^n}}}{{2\sqrt n }}\) converges and the value of limit is 0.

Most popular questions for Math Textbooks


Find the first 40 terms of the sequence defined by\({a_{n + 1}} = \left\{ {\begin{aligned}{\frac{1}{2}{a_n}}&{{\rm{ if }}{a_n}{\rm{ is an even number }}}\\{3{a_n} + 1}&{{\rm{ if }}{a_n}{\rm{ is an odd number }}}\end{aligned}} \right.;{a_1} = 11\).

Do the same if\({a_1} = 25\). Make a conjecture about this type of sequence

Determine whether the sequence converges or diverges. If it converges, find the limit.

\({a_n} = \frac{{{{\cos }^2}n}}{{{2^n}}}\)

Determine whether the series is convergent or divergent. If its convergent, find its sum.

\(\sum\limits_{n = 1}^\infty {\frac{{(1 + {3^n})}}{{{2^n}}}} \)

Show that the sequence defined by \({a_1} = 2,{a_{n + 1}} = \frac{1}{{3 - {a_n}}}\) sequences \(0 < {a_n} \le 2\) and is decreasing . Deduce that the sequence is covering and find its limit.

Determine whether the sequence converges or diverges. If it converges, find the limit.

\({\bf{\{ 0,1,0,0,1,0,0,0,1, \ldots \ldots \ldots \ldots \ldots \} }}\)
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