Geometric Series question with summations Calc 2 question.

Area of a equilateral triangle with sides of 1 is $\frac { \sqrt3}{2}$

You can find the radius of the largest circle by using pythagorean theorem.

Radius of the greatest circle is $\frac { 1 }{2\sqrt3}$ Making the area of the greater circle $\frac {\pi}{12}$

The radius of the next circle is $\frac { 1 }{4\sqrt3}$

The area is $\frac {\pi}{48}$ There are now three circle so the area of all three circles can simplify to $\frac {\pi}{16}$

The area of the next circle is $\frac {\pi}{192}$ or simplified to $\frac {\pi}{64}$ when adding the area of all three circles. The area of the circles are decreasing at $\pi \sum _{ 2 }^{ \infty }{ \frac { 1 }{ 4 } } ^{ n } = \pi\frac {1}{12}$. So the sum of the total area of the circles is $\frac { \pi }{ 12 } + \frac { \pi }{ 12 } = \frac {\pi}{6}$

I have googled this problem and others have received different results. I don't have a key to check my work and I'm curious what others have received and where is my mistake. First note that the area of the equilateral triangle is $\frac{\sqrt{3}}4$ not $\frac{\sqrt{3}}2$. Also note that the radius of the inner circle in equilateral triangle is one-third of its altitude. Meaning that $\frac 23$ of the altitude $h$ is reserved by the big circle and the remaining $\frac h3$ is assigned to the smaller circles in a similar way. Let $S_1$ be the area of the big circle, $S_2$ the sum of the areas of the three smaller circles, and so on. We can write: \begin{align} S_1&=\pi\left({\frac h3}\right)^2\\ S_2&=3\times\pi\left({\frac 13\times\frac h3}\right)^2=\frac 13 S_1\\ S_3&=3\times\pi\left({\frac 13\times\frac 13\times\frac h3}\right)^2=\frac 1{27}S_1\\&\cdots \end{align} In conclusion: \begin{align} S=\sum_{n=1}^\infty S_n&=S_1+\frac 13 S_1\left(1+\frac 19+\frac 1{81}+\cdots\right)\\ &=S_1+\frac 13 S_1\left(\frac 98\right)=\frac{11}8 S_1\\ &=\frac{11}8\pi\left({\frac {\sqrt{3}}6}\right)^2=\frac{11}{96}\pi \end{align}