Geometric Series Explanation - Rhea

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 $, we know

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

and thus

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

Questions and comments

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