Endless series of rectangles

Algebra Level 2

An endless series of rectangles is constructed each with a width 1 1 and height 1 n 1 n + 1 \dfrac {1}{n} - \dfrac{1}{n+1} , where n = 1 , 2 , 3 , . . . n = 1, 2, 3, ... Find the total areas of rectangles.


The answer is 1.

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2 solutions

The area of the first k k such rectangles is

1 n = 1 k [ 1 n 1 n + 1 ] = 1 + ( 1 2 + 1 2 ) + ( 1 3 + 1 3 ) + . . . + 1 k 1 + ( 1 k + 1 k ) 1 k + 1 = 1 1 k + 1 \begin{aligned}1\cdot\sum_{n=1}^{k}\left [\frac{1}{n}-\frac{1}{n+1}\right ]&=1+\left (-\frac{1}{2}+\frac{1}{2}\right )+\left (-\frac{1}{3}+\frac{1}{3}\right )+...+\frac{1}{k-1}+\left (-\frac{1}{k}+\frac{1}{k}\right )-\frac{1}{k+1}\\&=1-\frac{1}{k+1}\end{aligned}

As k k\to\infty , 1 k + 1 0 \frac{1}{k+1}\to 0 , so the total area of the rectangles is 1 0 = 1 1-0=\color{#20A900}{\boxed{1}} .

Chew-Seong Cheong
Aug 11, 2020

The area is given by

A = 1 × ( 1 1 1 2 ) + 1 × ( 1 2 1 3 ) + 1 × ( 1 3 1 4 ) + = 1 1 2 + 1 2 1 3 + 1 3 1 4 + 1 4 = 1 \begin{aligned} A & = 1 \times \left(\frac 11 - \frac 12 \right) + 1 \times \left(\frac 12 - \frac 13 \right) + 1 \times \left(\frac 13 - \frac 14 \right) + \cdots \\ & = 1 - \red{\cancel{\frac 12}} + \blue{\cancel{ \frac 12}} - \red{\cancel{\frac 13}} + \blue{\cancel{ \frac 13}} - \red{\cancel{\frac 14}} + \blue{\cancel{ \frac 14}} - \cdots \\ & = \boxed 1 \end{aligned}

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