## Geometric SeriesGeometric series are often one of the first examples one sees of the idea of an infinite sum converging to a finite number, and in fact they are simple enough that we can even derive a formula for the general case
$$\sum_{n=1}^\infty \frac{1}{r^n}=\frac{1}{r-1}$$ Here we will first look at a few specific examples where $r$ is a small integer, in order to gain intuition for what summing infinitely many things means. |

## $$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$$ |
The first infinite series that we should think about is the sum of the reciprocals of the powers of 2. Writing out the first couple of terms of this infinite sum we see
$$\sum_{n=1}^\infty\frac{1}{2^n}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}+\frac{1}{1024}\cdots$$ Clearly, the successive terms are dropping in size, and quickly at that! But why should we believe that the entire sum converges to some finite value? Remember, the sum $1+\frac{1}{2}+\frac{1}{3}+\cdots$ does not converge, even though the terms get smaller and smaller forever.Well, let's start out by trying to draw a picture of our sum. I'm going to start by drawing a square of area $1$, and for each term in the sum I will try to color in a chunk of the square with that area. For example here are my pictures for the first three partial sums: From here, it should be clear that the sum does not escape to infinity. Each time we need to add a new term, we just use half of the
remaining white space of the square. All of our partial sums are represented by the blue portion of the square, and since the total square area is 1, the blue area can never exceed 1. So, we have a definite upper bound on the value of our sum,$$\sum_{n=1}^\infty\frac{1}{2^n}\leq 1$$ Can we use this pictorial technique to calculate its exact value? Let's keep going and see! To add the next terms in the series, we just need to continue shading in half of the remaining whitespace. Where we have continually cut the white space over and over, adding it to our sum. As this process continues forever, the amount of whitespace tends to zero, and the amount of blue to 1. That is, the value of our sum tends to 1. That means we've found the sum of our series!
$$\sum_{n=1}^\infty \frac{1}{2^n}=1$$ |

## $$\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\cdots$$ |
Let's now turn our thoughts to a similar series, namely the sum of the recriprocals of powers of 4. We can ask ourselves the same two questions about this series: namely, does it converge to some finite value, and if so which value? It just so happens we can answer both of them with a similar process to the above. Lets begin again with the unit square, and to represent the first term of this series (1/4), lets shade in a quarter of it green. The next term is (1/4)2, or a fourth of one forth. We can represent this term by taking one of the remaining white squares, and coloring in a quarter of it, and similarly for the third term.
In the limit, as we allow the number of terms to grow without bound, we keep doing this forever and get a diagonal of smaller and smaller green squares. Like before, we can see that all of the green squares are contained within the original square, and so the value of this sum is less than 1.
To calculate its exact value this time though we are going to have to be a little more clever. First, notice that directly to the right of every green square is a white square of exactly the same area, and so we can slide each green square over "one to the right" to get another collection of squares of the same area. Likewise, there is a square above each green square of the same area so we can slide them "up one" as well.
But thats kind of amazing! We have found three copies of our sequence of green squares (whose total area represents the value of the infinite sum) inside of the original square. Let's draw them in there all at once.
This fills the entire square! (More precisely, at stage $n$ the uncolored area is $1/4^n$ which goes to $0$ as $n\to\infty$). As these three collections of squares share no overlap, this means that three times the area of our sequence of squares is the area of the original square (which was 1). And so, the area of our sequence must be $1/3$!
$$\sum_{n=1}^\infty \frac{1}{4^n}=\frac{1}{3}$$ |

## $$\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\cdots$$ |
Before trying to tackle the general case, let's do one more example. If we want to sum the reciprocal powers of 3 we could start by dividing a square into thirds, and then successively dividing one of those thirds into thirds...
Now, notice that instead of coloring the leftmost rectangle each time I could have instead colored the rightmost rectangle, and so fit two copies of my sequence of areas inside of the square.
And again, it turns out that these two copies fill up the original square completely, and without overlap (the uncolored portion of the square at step $n$ has area $1/3^n$ which is going to $0$ as $n\to\infty$. Thus, twice the sum of our series is $1$ and so
$$\sum_{n=1}^\infty \frac{1}{3^n}=\frac{1}{2}$$ |

## Some challenge questions |
While its fun to find clever ways to subdivide a square to calculate the sums of particular series, this doesn't seem to lend itself all that well to finding a general procedure - indeed all three ways above used totally different ways of breaking the square up!
Before worrying about generalities though here are a couple challenges: Can you find a way to sum the series $\frac{1}{8}+\frac{1}{8^2}+\cdots$ by a clever subdivision of a cube? How about an alternative argument that $\frac{1}{4}+\frac{1}{4^2}+\cdots=\frac{1}{3}$ by subdividing an equilateral triangle instead of a square? |