Question

Infinite rectangles each of width 1 unit and height $\left(\frac{1}{n}-\frac{1}{n+1}\right)(n \in N)$ are constructed such that ends of exactly one diagonal of every rectangle lies along the curve $y=\frac{1}{x}$. The sum of areas of all such rectangles, is

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{3}{4}$
• D. 1

Answer 1

Answer: (D)

Method-I:
$\text { Required sum }=\underset{n \rightarrow \infty}{\text{Lim}} \left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots \ldots+\left(\frac{1}{n}-\frac{1}{n+1}\right)\right]=\underset{n \rightarrow \infty}{\text{Lim}}\left(1-\frac{1}{n+1}\right)=1$
Method-II:

$\text { Area of } r ^{\text {th }} \text { rectangle } A_r=(\text { height }) \times(\text { width })=\left(\frac{1}{r}-\frac{1}{r+1}\right) \cdot 1 $
$\text { Sum of areas }=\underset{n \rightarrow \infty}{\text{Lim}} \displaystyle\sum_{r=1}^n A_r=\operatorname{Lim}_{n \rightarrow \infty}\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots . .+\left(\frac{1}{n}-\frac{1}{n+1}\right)\right]$
$=\underset{n \rightarrow \infty}{\text{Lim}}\left(1-\frac{1}{n+1}\right)=1$