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

$\text{ Area of }{r}^{\text{th }}\text{ rectangle }{A}_r=(\text{ height })\times (\text{ width })=\left(\frac{1}{r}-\frac{1}{r+1}\right)\cdot 1$
$\text{ Sum of areas }=\underset{n\to \infty }{\mathrm{Lim}}\sum _{r=1}^n{A}_r={\mathrm{Lim}}_{n\to \infty }\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots ..+\left(\frac{1}{n}-\frac{1}{n+1}\right)\right]$
$=\underset{n\to \infty }{\mathrm{Lim}}\left(1-\frac{1}{n+1}\right)=1$