The Geometric Series
(1/(1 - x)) = 1 + x + (x^2) + (x^3) + (x^4) + ⋯

by
K. D.  Stroyan
University of Iowa

copyright 1997 by Academic Press, Inc. - All rights reserved.

Special Mathematica Functions used in this NoteBook

A Function to Plot Powers: plotPowers[n,epsilon]

Off[General::spell1];
Needs["Graphics`Colors`"];
plotPowers[h_,epsilon_]:= 
Module[{functions,graylevels,curves,smallScale,combined,n,i,j,x},
Do[
	functions = Table[x^i, {i, n, 1, -1}] ;
	graylevels = Table[{GrayLevel[j]}, {j, 0, 0.5, 0.5/n}] ;
curves = Plot[Release[functions], {x, -1.1, 1.1},
		PlotStyle -> graylevels,DisplayFunction->Identity];
smallScale = Show[curves,PlotRange->{-epsilon,epsilon},Axes->None,
		AspectRatio->1/20,PlotLabel->BelowEpsilon,DisplayFunction->Identity];		
combined = Show[{curves,
	Graphics[{Red,AbsoluteThickness[5],
	Line[{{-epsilon^(1/n),0},{epsilon^(1/n),0}}]}],
	Graphics[{Red,AbsolutePointSize[5],
	Point[{1,1}]}],
	Graphics[{Orange,AbsolutePointSize[5],
	Point[{-1,(-1)^n}]}]
	},
	PlotRange -> {-1.1, 1.1},AspectRatio -> 1,DisplayFunction->Identity];

Show[GraphicsArray[{combined,smallScale}],
	DisplayFunction->$DisplayFunction];
,{n, 2, h}] ];

seriesGraphs[.]

Needs["Graphics`Colors`"];
seriesGraphs[f_, series_, radius_, numterms_Integer, opns___Rule] :=
	Module[ {functions, graylevels, i, j,n,x,terms},
		Do[
			functions = Table[series[x,n], {n, 0, i}] ;
			AppendTo[functions, f[x]] ;
			graylevels = Table[{GrayLevel[j]}, {j, 6/10, 0, -6/(10 i)}] ; 
			graylevels=AppendTo[graylevels,{AbsoluteThickness[2],Red}];
Plot[Release[functions], {x, -radius, radius},
		PlotStyle -> graylevels,
		PlotLabel ->i"terms", opns,
		AspectRatio->Automatic];
	,{i, 1, numterms}];
	Do[
	Plot[Release[{f[x], series[x,n]}],{x, -radius, radius},
	PlotLabel ->n"terms ", opns,
	PlotStyle -> {{AbsoluteThickness[2],Red},{GrayLevel[0]}},AspectRatio->Automatic];
	,{n, 1, numterms}];
	];

Notebook Overview

This NoteBook analyses the convergence of the geometric series in detail:

When | x | <= r < 1, we have  1 + x + (x^2) + (x^3) + ⋯ + (x^n) -> 
        (1/(1 - x)), as  n->∞

When  x = -1, 1 + x + (x^2) + (x^3) + ⋯ + (x^n)  oscillates endlessly:1,0,1,0,1,0,....

When x ≥ 1,  1 + x + (x^2) + (x^3) + ⋯ + (x^n)  grows larger and larger, or tends to +Infinity

The analysis requires only simple algebra and a clear understanding of the FUNCTION limit

Limit[(x^n), n -> ∞]

We begin with that limit.

References to the Text

This NoteBook goes with Chapter 25 of the text, but convergence of the geometric series is the foundation result used in proving the ratio and root tests for convergence of power series in Chapter 26.

Problem 25.1

This whole NoteBook is Problem 25.1.  It shows you how the geometric series converges.  Answer the questions as you go to solve the exercise.

Convergence of (x^n)

The functions  y = x,  y = (x^2),  y = (x^3),  y = (x^4),  y = (x^5), 
      ⋯ tend to zero inside any interval of the form  -r ≤ x ≤ r, when  0 < r < 1.  Specifically, if we control the maximum absolute size of  x ,  | x | ≤ r, then we can find an  m  large enough so that  | x^n | < ϵ  for all  n>m and all  x  with  -r ≤ x ≤ r.  This isn't really very hard once we "see" what is going on.

High-Tech Proof

Here's the high-tech mathematical proof that
	(x^n) ->  0
uniformly inside -r ≤ x ≤ r, for fixed r, 0 < r < 1:

Fix  r, 0 < r < 1, and fix  epsilon > 0.  
Solve for  m:

(r^m) = ϵLog[(r^m)] = Log[ϵ]m Log[r] = Log[ϵ]m = Log[ϵ]/Log[r]

Then whenever | x | <= r and n > m,  | x^n | <ϵ.

This proof is compact, simple in itself, but hard to understand.  There is too much "legal fine print" at a first look.

More Dynamic Approach

If we don't control  x  with the maximum  r < 1  in advance, there is another way to describe what is happening with the function limit of  (x^n).  Any value of  x  with  | x | < 1  has  | (x^n) |  getting smaller and smaller as  n  gets larger and larger.  This is simply because  (x^n) = x * (x^(n - 1)), is a multiple of a number less than  1  with another number, so  | (x^n) | < | (x^(n - 1)) |.

Consider choosing a very high power (like m = 20).  If we want our limit approximation  x^m  to be within, say  0.01 of zero, by looking at the graphs (or knowing this in advance or using calculus) we know that the biggest value of  x^n  will be at the highest value of  x.  Let's find the maximal  r that still has r^n < 0.01,

(r^m) = 0.01r = (0.01^(1/m))

or in general

(r^n) = ϵr = (ϵ^(1/m))

Then we know that for  -r ≤ x ≤ r and n>m, | (x^n) | < ϵ.  This approach shows us how much of our approximation is "good."  It doesn't really resolve the "legal" difficulties in the high-tech proof, because  r  moves (and in fact, Limit[(ϵ^(1/m)), m -> ∞]= 1.)  It is easier to "see."

20 Graphs

Let's see for ourselves how this approach looks by <Enter>ing th next computation.  Press <Command> & <y> after highlighting the cell with the graphs of  y = x & y = (x^2) to see successive plots of  y = (x^n)  for n = 1 to 20...

epsilon = 0.01; Print["Red bar within epsilon = ", epsilon]; plotPowers[20, epsilon]

Practice Exercise

Compute in advance how much of  x^n  for  n>30  lies within 0.01 of the final limit, 0.  In other words, find  r so that if  -r <= x <= r, then  | x^n | < 0.01.

epsilon = 0.01;
r = ???
plotPowers[30,epsilon]

Your Description of Theorem 25.2

Explain in your own words what the complete limiting behavior of the functions
y = x,  y = (x^2),  y = (x^3),  y = (x^4),  y = (x^5), 
      ⋯ is as  n -> Infinity:

1)  For fixed  r, 0 < r < 1, when -r <= x <= r:
YOUR DESCRIPTION:

2)  For  x >= 1:
YOUR DESCRIPTION:

3)  For fixed  x =-1:
YOUR DESCRIPTION:

4)  For  x < -1:
YOUR DESCRIPTION:

5)  Non-convergence near the boundary:
For every power  n, no matter how large, there is always a point  x  with  |x| < 1  so that  x^n  is at least  1/2 unit away from the ultimate limit of  x^n.  (Notice the discontinuity in the limit.)  Why is this?
YOUR DESCRIPTION:

Step-by-step Geometric Convergence: Theorem 25.1

The Algebra of  1 + x + (x^2) + (x^3) + ⋯ + (x^n)

Show the identity

((1 - (x^(n + 1)))/(1 - x)) =  
        1 + x + (x^2) + (x^3) + ⋯ + (x^n)  (x ≠ 1)

Here is some Mathematica help:

Clear[x, k, n]; (1 - x) (∑_(k = 0)^n) (x^k)

n = 10; (1 - x) (∑_(k = 0)^n) (x^k)Expand[(1 - x) (∑_(k = 0)^n) (x^k)]

n = 100; Expand[(1 - x) (∑_(k = 0)^n) (x^k)]

n = 1000; Expand[(1 - x) (∑_(k = 0)^n) (x^k)]

The Limit of (1/(1 - x)) - ((x^(n + 1))/(1 - x))

Use convergence of  (x^n)  on an interval  -r <= x <= r (described in you solution to the first section) to show that for r fixed, 0 < r < 1,

1 + x + (x^2) + (x^3) + ⋯ + (x^n) -> 
        (1/(1 - x))

for  -r ≤ x ≤ r .  <Enter> the next animation now and expalin it below.

f[x_] := (1/(1 - x)); series[x_, n_] := (∑_(k = 0)^n) (x^k); radius = 1.2; numterms = 10; seriesGraphs[f, series, radius, numterms, 
      PlotRange -> {-1, 4}];

Your analysis of the convergence of the geometric series

Explain the above animation in terms of your analysis of the limit.  Include:
The good news for  -r ≤ x ≤ r:
YOUR ANALYSIS:
  
The wiggling for  x≤-1:
YOUR ANALYSIS:
  
The lack of graphs for  x≥1:
YOUR ANALYSIS: