Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

Show that the series is convergent. How many terms of the series do we need to add in order to find the sum to the indicated accuracy?
\(\sum\limits_{n = 0}^\infty {{{( - 1)}^{n - 1}}n{e^{ - n}}{\rm{ (}}|{\rm{error}}|{\rm{ }} < 0.01)} \)

Yes, the series is convergent and the number of terms we need to add atleast \(6\) terms.

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TABLE OF CONTENTS :

TABLE OF CONTENTS

Alternating series test:

\(n{e^{ - n}}\) will always be positive for positive \(n\) and \({( - 1)^{n - 1}}\) is positive on odd terms and negative on even terms.

Also, the terms decrease in size. For any \(k\),consider the size difference between the \({(k + 1)^{{\rm{th}}}}\) and \({k^{{\rm{th}}}}\) term

\(\begin{aligned}(k + 1){e^{ - (n + 1)}} - n{e^{ - n}} &= \frac{{(n + 1)}}{{{e^{n + 1}}}} - \frac{{ - n}}{{{e^n}}}\\ &= \frac{{{e^n}(n + 1) - n{e^{n + 1}}}}{{{e^{2n + 1}}}}\\ &= \frac{{{e^n}\left( {n(1 - e) + 1} \right)}}{{{e^{2n + 1}}}}\end{aligned}\)

\(1 - e < - 1\) and \(n \ge 1\)

So, \(\begin{aligned}n(1 - e) < - 1\\n(1 - e) + 1 < 0\end{aligned}\)

So the terms are decreasing in size .

Checking the limit

\(\mathop {\lim }\limits_{n \to \infty } {\rm{ }}n{e^{ - n}} = \mathop {\lim }\limits_{n \to \infty } {\rm{ }}\frac{n}{{{e^n}}}\)

If \(\mathop {\lim }\limits_{n \to \infty } {\rm{ }}\frac{n}{{{e^n}}}\) exists,

Let \(x = n\), then \(\mathop {\lim }\limits_{x \to \infty } {\rm{ }}\frac{x}{{{e^x}}} = \mathop {\lim }\limits_{n \to \infty } {\rm{ }}\frac{n}{{{e^n}}}\)

Then \(\begin{aligned} \mathop {\lim }\limits_{x \to \infty } {\rm{ }}\frac{x}{{{e^x}}} &= \mathop {\lim }\limits_{n \to \infty } {\rm{ }}\frac{1}{{{e^x}}}\\\mathop {\lim }\limits_{x \to \infty } {\rm{ }}\frac{x}{{{e^x}}} &= 0\end{aligned}\)

Therefore, \(\mathop {\lim }\limits_{n \to \infty } {\rm{ }}\frac{n}{{{e^n}}} = 0\)

So, the series is converging.

Let b_nbe the sequence \(n{e^{ - n}}\), \(S = \sum\limits_{n = 1}^\infty {{{( - 1)}^{n - 1}}n{e^{ - n}}} \)

Then \(|S - {S_n}| \le {b_{n + 1}}\)

We want \(n\) with \({b_{n + 1}} < 0.01\)

So that

\(\begin{aligned} (n + 1){e^{ - (n + 1)}} < 0.01\\\frac{{n + 1}}{{0.01}} < {e^{n + 1}}\end{aligned}\)

Note that \(\frac{{{e^6}}}{6} = 67.2 < 100\), while \(\frac{{{e^7}}}{7} = 156.6 < 100\)

So, \(6\) is the least number that works for \(n + 1\)

Therefore, we need to add atleast \(6\) terms

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Essential Calculus: Early Transcendentals

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Short Answer

Show that the series is convergent. How many terms of the series do we need to add in order to find the sum to the indicated accuracy?
\(\sum\limits_{n = 0}^\infty {{{( - 1)}^{n - 1}}n{e^{ - n}}{\rm{ (}}|{\rm{error}}|{\rm{ }} < 0.01)} \)

Step by Step Solution

Alternating series test:

Checking the limit

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Show that the series is convergent. How many terms of the series do we need to add in order to find the sum to the indicated accuracy?\(\sum\limits_{n = 0}^\infty {{{( - 1)}^{n - 1}}n{e^{ - n}}{\rm{ (}}|{\rm{error}}|{\rm{ }} < 0.01)} \)

Step by Step Solution

Alternating series test:

Checking the limit

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Show that the series is convergent. How many terms of the series do we need to add in order to find the sum to the indicated accuracy?
\(\sum\limits_{n = 0}^\infty {{{( - 1)}^{n - 1}}n{e^{ - n}}{\rm{ (}}|{\rm{error}}|{\rm{ }} < 0.01)} \)