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Essential Calculus: Early Transcendentals
Found in: Page 452
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

Draw a picture to show that

\(\sum\limits_{n = 2}^\infty {\frac{1}{{\mathop n\nolimits^{1.3} }}} < \int_1^\infty {\frac{1}{{\mathop x\nolimits^{1.3} }}} dx\)

What can you conclude about the series?

\(\sum\limits_{n = 2}^\infty {\frac{1}{{\mathop n\nolimits^{1.3} }}} \) is conversion

See the step by step solution

Step by Step Solution

With the help of making a graph from equation,

Proof :

The given series is represented by the sum of the area of rectangles.

We can conclude from the figure that the area under the curve is less than the area of the rectangle. the curve is

\(y = \frac{1}{{\mathop x\nolimits^{13} }}\) for \(x \ge 1\), Which is the value of \(\int_1^\infty {\frac{1}{{\mathop x\nolimits^{1.3} }}} dx\)

Therefore, \(\sum\limits_{n = 2}^\infty {\frac{1}{{\mathop n\nolimits^{1.3} }}} < \int_1^\infty {\frac{1}{{\mathop x\nolimits^{1.3} }}} dx\)

Now that the integral \(\int_1^\infty {\frac{1}{{\mathop x\nolimits^{1.3} }}} dx\) converges, so by the integral test, \(\int_1^\infty {\frac{1}{{\mathop x\nolimits^{1.3} }}} dx\) also converges

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