Interpreting images representing geometric series I understand the formula for infinite geometric series as 
$$S = \frac{a_{1}}{1-r}$$ if $0<r<1$
However I'm having trouble applying it to these images 
It seems to me that in the first image, the first square represents 1/4 of the entire square
For the second and third images, the respective rectangle and triangle make up 1/2 of the entire square. 
Not sure what to do with this. Does it mean that for the first image for example, the image is $$\sum_{n=0}^\infty \left(\frac{1}{4}\right)^n$$

1) The area of the first shaded square is a fourth part of the original square: $\frac{1}{4}$. The area of the second shaded square is a fourth of a fourth of the original area: $\frac{1}{4}\cdot\frac{1}{4}$. The area of the third square would be a fourth of that: $\frac{1}{4}\cdot\frac{1}{4}\cdot\frac{1}{4}$. Do you see the pattern?
$$
\frac{1}{4}+\left(\frac{1}{4}\cdot\frac{1}{4}\right)+\left(\frac{1}{4}\cdot\frac{1}{4}\cdot\frac{1}{4}\right)+...=\\
\left(\frac{1}{4}\right)^1+\left(\frac{1}{4}\right)^2+\left(\frac{1}{4}\right)^3+...=\\
\sum_{n=1}^{\infty}\left(\frac{1}{4}\right)^n=
\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^n-1=
\frac{1}{1-\frac{1}{4}}-1=\frac{4}{3}-1=\frac{1}{3}.
$$
2) The first rectangle is area $\frac{1}{2}$. The second is a half of the original area divided by four $\frac{1}{2}\cdot\frac{1}{4}$. The third part is one fourt of that $\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{4}$:
$$
\frac{1}{2}+\left(\frac{1}{2}\cdot\frac{1}{4}\right)+\left(\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{4}\right)+...=\\
\frac{1}{2}\left(\frac{1}{4}\right)^0+\frac{1}{2}\left(\frac{1}{4}\right)^1+
\frac{1}{2}\left(\frac{1}{4}\right)^2+...=\\
\sum_{n=0}^{\infty}\frac{1}{2}\left(\frac{1}{4}\right)^n=\frac{1}{2}\cdot\frac{1}{1-\frac{1}{4}}=\frac{2}{3}.
$$
3) The first triangle is area $\frac{1}{2}$. The second triangle is area $\frac{1}{2}\cdot\frac{1}{4}$ (a fourth part of a half of the original triangle). The third triangle is going to have area $\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{4}$. I think you see that the pattern is the same as in the previous case:
$$
\frac{1}{2}+\left(\frac{1}{2}\cdot\frac{1}{4}\right)+\left(\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{4}\right)+...=\\
\frac{1}{2}\left(\frac{1}{4}\right)^0+\frac{1}{2}\left(\frac{1}{4}\right)^1+
\frac{1}{2}\left(\frac{1}{4}\right)^2+...=\\
\sum_{n=0}^{\infty}\frac{1}{2}\left(\frac{1}{4}\right)^n=\frac{1}{2}\cdot\frac{1}{1-\frac{1}{4}}=\frac{2}{3}.
$$
