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:

$ a+ax+ax^2+ax^3+ax^4+\cdots=\frac{a}{1-x} $ if $ |x|<1 $

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+x+x^2+\cdots+x^n $

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

$ S_n - xS_n = (1+x+x^2+\cdots+x^n) - x(1+x+x^2+\cdots+x^n) $

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

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

$ S_n(1-x) = 1-x^{n+1} $

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

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

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

$ lim_{n\rightarrow\infty} \frac{x^{n+1}}{1-x} = 0 $.

One can see this by example. Imagine, for example that x is .9. Then if we take .9 / (1 - .9) and multiply it repeatedly by .9, it will get as small as we wish (just multiply it by .9 enough times). This will work for any number between -1 and 1. Once you believe this, we get the result:

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

This is great! This isn't quite the formula I listed above, but you can see that

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

and

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

Examples, for algebraic practice

Example 1

Questions and comments

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