Question

A square in inscribed in a circle of radius R, a circle is inscribed in the square, a new square in the circle and so on for 'n' times.

Find the limit of sum of areas of all the squares as $n\to \infty$.

Answer 1

We have,

(Refer image)

The area of the first square can be

calculated if side is known.

So, we have length of diagonal

$2R=\sqrt{2}a$ (a $=$ side length)

$\Rightarrow a=\sqrt{2}R$

and the radius of next circle

will be $\frac{a}{2}=\frac{R}{\sqrt{2}}$

and So on

thus the side length of every

inscribed square will be , $\frac{1}{\sqrt{2}}$ times

the previous square's side length

So, the area will be $\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}$ times

the previous square's area so

the sum will be,

$S_n=(\sqrt{2}R{)}^2+(\frac{\sqrt{2}R}{2}{)}^2+(\frac{\sqrt{2}R}{4}{)}^2+......(\frac{\sqrt{2}R}{2^{n-1}})$

$\underset{n\to \infty }{lim}{S}_n=(\sqrt{2}R{)}^2\left[\frac{1}{1-\frac{1}{2}}\right]=2\times 2R^2=4R^2$