Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

Graph the curves \(y = {x^n}\), \(0 \le x \le 1\), for \(n = 0,1,2,3,4,....\)on a common screen. By finding the areas between successive curves, give a geometric demonstration of the fact, shown in Example 6, that
\(\sum\limits_{n = 1}^\infty {\frac{1}{{n(n + 1)}}} = 1\)

The area of the successive curve is

\(1 = \sum\limits_{n = 1}^\infty {\frac{1}{{n\left( {n + 1} \right)}}} \)

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Plotting the given function

First, we have to consider the curve,

\(y = {x^n}\), \(0 \le x \le 1\)

To find the areas between the successive curves, we have to plot the graphs of the given function to its limits.

The graph will be shown as

From the graph, it seems to be that the entire unit square \([0,1] \times [0,1]\) is eventually filled up by the regions that lie between successive functions.

Area of the region

Let’s take an example, the region between \(x\) and \({x^2}\) is shaded below

Similarly, the region between \({x^{n - 1}}\) and \({x^n}\) for all \(n\).

These regions fill up the entire square. This can be proved using an argument with limits, but it is clear from the pictures that this eventually happens.

Now, the area of the region between \({x^{n - 1}}\) and \({x^n}{a_n}\). Then compute \({a_n}\) using integration

\(\begin{aligned}{a_n} &= \int\limits_0^1 {{x^{n - 1}}} - {x^n}\\{a_n} &= \left[ {\frac{1}{n}{x^n} - \frac{1}{{n + 1}}{x^{n + 1}}} \right]_0^1\\{a_n} &= \frac{1}{n} - \frac{1}{{n + 1}}\\{a_n} &= \frac{{n + 1}}{{n\left( {n + 1} \right)}} - \frac{n}{{n\left( {n + 1} \right)}}\\{a_n} &= \frac{1}{{n\left( {n + 1} \right)}}\end{aligned}\)

And, here we observed that the connection to the original series.

The unit square is, as noted, the union of all of these regions, and none of them overlap.

Therefore, the area of the unit square is equal to the sum of the area of the regions. But the area of the unit square is \(1\), and the \({n^{th}}\)region has area \({a_n}\).

Therefore, the area of the successive curve is

\(\begin{aligned}1 &= \sum\limits_{n = 1}^\infty {{a_n}} \\1 &= \sum\limits_{n = 1}^\infty {\frac{1}{{n\left( {n + 1} \right)}}} \end{aligned}\)

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Graph the curves \(y = {x^n}\), \(0 \le x \le 1\), for \(n = 0,1,2,3,4,....\)on a common screen. By finding the areas between successive curves, give a geometric demonstration of the fact, shown in Example 6, that
\(\sum\limits_{n = 1}^\infty {\frac{1}{{n(n + 1)}}} = 1\)

Step by Step Solution

Plotting the given function

Area of the region

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Graph the curves \(y = {x^n}\), \(0 \le x \le 1\), for \(n = 0,1,2,3,4,....\)on a common screen. By finding the areas between successive curves, give a geometric demonstration of the fact, shown in Example 6, that \(\sum\limits_{n = 1}^\infty {\frac{1}{{n(n + 1)}}} = 1\)

Step by Step Solution

Plotting the given function

Area of the region

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Graph the curves \(y = {x^n}\), \(0 \le x \le 1\), for \(n = 0,1,2,3,4,....\)on a common screen. By finding the areas between successive curves, give a geometric demonstration of the fact, shown in Example 6, that
\(\sum\limits_{n = 1}^\infty {\frac{1}{{n(n + 1)}}} = 1\)