How to show graphically that $\sum_i(1/3)^i$ goes to $1/2$ As everybody knows, it is very easy to show:
$$\sum_{i=1}^\infty \frac{1}{2^i} = 1$$
As follows:

The coloured parts always show $\frac{1}{2^i}$ and it's easy to see they all come together to fill the entire square.
Does anybody know a similar drawing to show that:
$$\sum_{i=1}^\infty \frac{1}{3^i} = \frac{1}{2}$$
Thanks in advance
 A: 
Start with any isosceles triangle of area $1$. Connect all its corners to a point such that we have three triangles of the same area. Repeat for successive isosceles triangles, all sharing the same base.
The yellow area is $\frac 13$, green area is $\frac 13 \times \frac 13 =\frac 19$, red area is $\frac 19 \times \frac 13 =\frac{1}{27}$ and so on.
The total area approaches exactly $\frac 12$.
A: This is a visual proof for
$$\sum_{i=1}^\infty \frac{1}{3^i} = \frac{1}{2}.$$
For any positive integer $i$:

*

*the term  $\frac{1}{3^{2i-1}}$ is given by the area of a rectangle $\frac{1}{3^{i}}\times \frac{1}{3^{i-1}}$.


*the term $\frac{1}{3^{2i}}$ is given by the area of a square $\frac{1}{3^{i}}\times \frac{1}{3^{i}}$.
Such squares and rectangles cover half of the square $1\times 1$.

I found the picture HERE
A: I like to use rectangles that I divide up based on the common ratio in the geometric series and compare filled to unfilled areas. This works well if the common ratio is less than $\displaystyle \frac{1}{2}$. I haven't yet figured out how to make it work for bigger than $\displaystyle \frac{1}{2}$.
The idea is to compare the number of filled rectangles and the unfilled rectangles of the same size (neglecting any rectangles that need to be split up for the next piece in the geometric series).
Here is a common ratio of $\displaystyle \frac{1}{3}$:

For $\displaystyle \frac{1}{3}$, there is one filled rectangle for every unfilled of the same size, meaning the areas are in a $1:1$ ratio. If the original rectangle's area is $1$, that must mean the filled areas are $\displaystyle \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots = \frac{1}{2}$ (in agreement with the geometric series formula).
Here is a common ratio of $\displaystyle \frac{1}{4}$:

For $\displaystyle \frac{1}{4}$, there is one filled rectangle for every  two unfilled rectangles of the same size, meaning the areas are in a $1:2$ ratio. If the original rectangle's area is $1$, that must mean the filled areas are $\displaystyle \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \cdots = \frac{1}{3}$ (in agreement with the geometric series formula).
Here is a common ratio of $\displaystyle \frac{2}{9}$:

For $\displaystyle \frac{2}{9}$, there are two filled rectangles for every  five unfilled rectangles of the same size, meaning the areas are in a $2:5$ ratio. If the original rectangle's area is $1$, that must mean the filled areas are $\displaystyle \frac{2}{9} + \frac{4}{81} + \frac{8}{729} + \cdots = \frac{2}{7}$ (in agreement with the geometric series formula).
Edit: After thinking for a few minutes, I realized how to do the case of the common ratio being bigger than $\displaystyle \frac{1}{2}$: multiply the common ratio by itself until you end up with a ratio less than $\displaystyle \frac{1}{2}$, then do the above process and add the portion that was chopped off at the end. I came up with this idea Summer of 2022 and after a lot of thought could not come up with this. Funny the difference some time makes.
