A square is inscribed in a circle of radius `R`, a circle is inscribed in this square then a square in this circle and so on `n` times. Find the limit of the sum of areas of all the squares as `n to oo`.

Answer

To solve the problem step by step, we will analyze the geometric relationships between the squares and circles, and then calculate the limit of the sum of the areas of the squares as \( n \) approaches infinity.

### Step 1: Determine the side length of the first square
Given a circle with radius \( R \), the diameter of the circle is \( 2R \). The diagonal of the inscribed square is equal to the diameter of the circle.

The diagonal \( d \) of a square with side length \( a \) is given by:
\[ d = a\sqrt{2} \]
Setting the diagonal equal to the diameter of the circle, we have:
\[ a\sqrt{2} = 2R \]
Solving for \( a \):
\[ a = \frac{2R}{\sqrt{2}} = R\sqrt{2} \]

### Step 2: Determine the side length of subsequent squares
The second square is inscribed in the circle that is inscribed in the first square. The side length of the second square is half that of the first square:
\[ a_2 = \frac{a_1}{\sqrt{2}} = \frac{R\sqrt{2}}{\sqrt{2}} = R \]

Continuing this pattern, the side lengths of the squares can be expressed as:
- First square: \( a_1 = R\sqrt{2} \)
- Second square: \( a_2 = R \)
- Third square: \( a_3 = \frac{R}{\sqrt{2}} \)
- Fourth square: \( a_4 = \frac{R}{2} \)
- And so on...

In general, the side length of the \( n \)-th square can be expressed as:
\[ a_n = \frac{R\sqrt{2}}{(\sqrt{2})^{n-1}} = R\frac{2^{(n-1)/2}}{2^{(n-1)/2}} = R\frac{1}{2^{(n-1)/2}} \]

### Step 3: Calculate the area of each square
The area \( A_n \) of the \( n \)-th square is given by:
\[ A_n = a_n^2 = \left(R\frac{1}{2^{(n-1)/2}}\right)^2 = R^2\frac{1}{2^{n-1}} = \frac{R^2}{2^{n-1}} \]

### Step 4: Sum the areas of all squares
The total area \( S_n \) of the first \( n \) squares is:
\[ S_n = A_1 + A_2 + A_3 + \ldots + A_n = R^2\left(1 + \frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{2^{n-1}}\right) \]

This series is a geometric series with the first term \( 1 \) and common ratio \( \frac{1}{2} \). The sum of the first \( n \) terms of a geometric series is given by:
\[ S_n = \frac{a(1 - r^n)}{1 - r} \]
where \( a \) is the first term and \( r \) is the common ratio. Thus:
\[ S_n = R^2 \cdot \frac{1(1 - (1/2)^n)}{1 - 1/2} = R^2 \cdot \frac{1 - (1/2)^n}{1/2} = 2R^2(1 - (1/2)^n) \]

### Step 5: Find the limit as \( n \to \infty \)
As \( n \) approaches infinity, \( (1/2)^n \) approaches \( 0 \). Therefore:
\[ \lim_{n \to \infty} S_n = 2R^2(1 - 0) = 2R^2 \cdot 2 = 4R^2 \]

### Final Result
The limit of the sum of the areas of all the squares as \( n \) approaches infinity is:
\[ \boxed{4R^2} \]