Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

(a) Use a graph of \(y = \frac{1}{x}\)to show that if\({S_n}\)is the \({n^{th}}\)partial sum of the harmonic series, then\({S_n} \le 1 + \ln n\).
(b) The harmonic series diverges, but very slowly. Use part(a) to show that the sum of the first million term is less than\(15\)and the sum of the first billion terms is less than\(22\).

(a) By using\(y = \frac{1}{x}\) graph, prove that\({S_n} \le 1 + \ln n\).

(b) BY using part(a) we proved that the sum of the first million term is less than\(15\)and the sum of the first billion terms is less than\(22\).

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Step by Step Solution

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TABLE OF CONTENTS

plotting graph using\(y = \frac{1}{x}\):

We take\(f\left( x \right) = \frac{1}{x}\)so\(\sum\limits_{n = 1}^\infty {\frac{1}{n}} \)is a harmonic series.

Prove that \({S_n} \le 1 + \ln n\):

From figure,

\(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ........... + \frac{1}{n} \le \int_1^n {\frac{1}{x}dx = \ln n} \).

So,

\({S_n} = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ........... + \frac{1}{n} \le 1 + \ln n\)

\({S_n} \le 1 + \ln n\).

Show that\({S_{10}}^9 < 22\)and\({S_{10}}^6 < 15\):

Since\({10^6} = 1million\) \({10^9} = 1million\)

If number of terms\(n = {10^6}\)then sum\({S_{10}}^6 \le 1 + \ln {10^6} \approx 14.82 < 15\).

So \({S_{10}}^6 < 15\).

If number of terms, \(n = {10^9}\).

The sum\({S_{10}}^9 \le 1 + \ln {10^9} \approx 21.72 < 22\).

So, \({S_{10}}^9 < 22\).

Therefore, (a) \({S_n} \le 1 + \ln n\)is proved.

b) We proved sum of the first million terms is less than\(15\) and sum of the first billion terms is less than\(22\).

That is, \({S_{10}}^6 < 15\)and\({S_{10}}^9 < 22\) hence proved.

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Essential Calculus: Early Transcendentals

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Step by Step Solution

plotting graph using\(y = \frac{1}{x}\):

Prove that \({S_n} \le 1 + \ln n\):

Show that\({S_{10}}^9 < 22\)and\({S_{10}}^6 < 15\):

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Step by Step Solution

plotting graph using\(y = \frac{1}{x}\):

Prove that \({S_n} \le 1 + \ln n\):

Show that\({S_{10}}^9 < 22\)and\({S_{10}}^6 < 15\):

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