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Essential Calculus: Early Transcendentals
Found in: Page 444
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

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

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

See the step by step solution

Step by Step Solution

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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