Difference between revisions of "Geometric Series Explanation" - Rhea

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Geometric Series

by Alec McGail, proud Member of the Math Squad.

INTRODUCTION A geometric series is a very useful infinite sum which seems to pop up everywhere:

If $ |r|<1 $,

$ a+ar+ar^2+ar^3+ar^4+\cdots=\frac{a}{1-r} $

Explanation

If you fully understand what the above statement means, you may skip this section. If you've ever worked with infinite sums, you probably have some idea of what the "$ \cdots $" is trying to imply. It's important, however, to know that mathematically, the expression

$ a_1 + a_2 + a_3 + \cdots $

is actually not a sum - at least not in the same sense as a finite sum.

Proof

There's a few different proofs I can think of for this fact, but there is one in particular which requires no extra "machinery", and is very convincing. First, define $ S_n $ to be the nth partial sum, i.e. the sum of the first n terms.

$ S_n=1+r+r^2+\cdots+r^n $

Then we can use a little trick to obtain a closed form expression for this sum:

$ S_n - rS_n = (1+r+r^2+\cdots+r^n) - r(1+r+r^2+\cdots+r^n) $

$ S_n(1-r) = 1+r+r^2+\cdots+r^n - (r+r^2+r^3+\cdots+r^{n+1}) $

Then we see that $ r $ and $ -r $ cancel, $ r^2 $ and $ -r^2 $ cancel, all the way up to $ r^n $ and $ -r^n $, leaving only

$ S_n(1-r) = 1-r^{n+1} = \frac{1-r^{n+1}}{1-r} $

$ S_n = \frac{1}{1-r} - \frac{r^{n+1}}{1-r} $

So then, if $ |r|<1 $, it's not terribly hard to see that

$ lim_{n\rightarrow\infty} S_n = \frac{1}{1-r} $

This is great! This isn't quite the formula I listed above, but

$ aS_n = a+ar+ar^2+\cdots+ar^n $

and

$ lim_{n\rightarrow\infty} aS_n = lim_{n\rightarrow\infty} a \left(\frac{1}{1-r} - \frac{r^{n+1}}{1-r}\right) = \frac{a}{1-r} $

So this is the result we were looking for.

Substitution

Examples

Example 1

Imagine John has a giant house-sized block of cheese he intends to share with the entire world. First he cuts the cheese in half and gives one half to his mom. He then cuts what's left in half and gives one piece to a friend. He continues this process indefinitely. In the limit, how much cheese will he have given away? Will he ever lose all of his cheese?

Both of these questions can be answered through intuition of situation alone, but we will formulate the question in terms of a geometric series. After n steps, the fraction of his cheese John will have given away is

$ S_n \;=\; \frac{1}{2} \,+\, \frac{1}{4} \,+\, \frac{1}{8} \,+\, \frac{1}{16} \,+\, \cdots \,+\, \frac{1}{2^n} $

because first he gives away half, then half of the half he has left (one fourth), etc. This is a geometric progression with common factor 1/2, and we need only apply the formula we just spend all that time talking about

$ \frac{1}{2} \,+\, \frac{1}{4} \,+\, \frac{1}{8} \,+\, \frac{1}{16} \,+\, \cdots = \frac{ 1/2 }{ 1 - 1/2 } = 1 $

Thus in the limit, he will have given away his entire cheese block.

Example 2

Exercises, for some practice

Exercise 1

$ \left(\frac{1}{4}\right)^3 \,+\, \left(\frac{1}{4}\right)^4 \,+\, \left(\frac{1}{4}\right)^5 \,+\, \left(\frac{1}{4}\right)^6 \,+\, \cdots $

Exercise 2

$ \sum_{n=0}^\infty \left( \frac{2^{n+1}}{5^n} \right) $

Exercise 3

$ \sum_{n=1}^\infty \left(-1\right)^n \frac{3}{4^n} $

Exercise 4

$ 1 + \frac{1}{\sqrt{2}} + \frac{1}{2} + \cdots $

Questions and comments

If you have any questions, comments, etc. please, please please post them below:

• Comment / question 1
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