02 824 365
Infinitia
Infinitia
Infinitia

Station  7B
Should we believe infinite sums? 

The geometric series formula is 1+x+x2+x3+...=11x1+x+x{2}+x{3}+... = \dfrac {1} {1-x}.

We saw this by taking the polynomial 11 and dividing it by the polynomial 1x1-x. The infinite sum of the powers of xx naturally appear.
I7S7B - Image034

But can this formula be true?

Question 1

What if xx is, say, the number 22?

Then the geometric series formula says that 1+2+4+8+16+...1+2+4+8+16+... equals 112\dfrac {1} {1-2}, which is 1-1. That’s just absurd!

On the other hand, if x=0.1x = 0.1, then the geometric series formula says that

1+(0.1)+(0.1)2+(0.1)3+...=1+0.1+0.01+0.001+...=1.111...1+\left(0.1\right)+\left(0.1\right){2}+\left(0.1\right){3}+... = 1+0.1+0.01+0.001+...=1.111...

equals

110.1=10.9=109\dfrac {1} {1-0.1}=\dfrac {1} {0.9}=\dfrac {10} {9}.

This is one and one-ninth. From primary school we learn that 19\dfrac {1} {9} as a decimal is 0.111...0.111.... (Also see the Decimalia Island!)
So one-and-one-ninth equals 1.111...1.111.... The geometric series formula is correct in this case!

1+(0.1)+(0.1)2+(0.1)3+...=110.11+\left(0.1\right)+\left(0.1\right){2}+\left(0.1\right){3}+... = \dfrac {1} {1-0.1}

What is going on? When can we believe the formula and when can we not?

Question 2

ALGEBRA VERSUS ARITHMETIC

The 1x1\leftarrow x machine is a machine that displays mechanical algebra. Everything it does and all we conclude from it is true in terms of mechanical algebra.

For example, the geometric series formula says that 1+2+4+8+...1+2+4+8+... equals 112\dfrac{1}{1-2}, which is absurd as a statement of arithmetic. However, in a purely mechanical sense, without regard to arithmetic, there is some version of truth to this claim. Here’s how we can see some truth.

If 1+2+4+8+...1+2+4+8+... equals 112\dfrac{1}{1-2}, then multiplying 1+2+4+8+...1+2+4+8+... by 121-2 should give 11. Does it?
Yes it does!

(12)×(1+2+4+8+...)=(12)+(12)×2+(12)×4+(12)×8+...=12+24+48+816+...=1+0+0+0+...=1\begin{aligned}
\left(1-2\right) \times \left(1+2+4+8+...\right) &= \left(1-2\right) + \left(1-2\right) \times 2 + \left(1-2\right) \times 4 + \left(1-2\right) \times 8 + ...\
&= 1-2+2-4+4-8+8-16+...\
&= 1+0+0+0+...\
&= 1
\end{aligned}

So 1+2+4+8+...1+2+4+8+... really does behave like 112\dfrac{1}{1-2}.
But still, this is not very satisfying. We want to know when 1+x+x2+x3+...=11x1+x+x{2}+x{3}+... = \dfrac {1} {1-x} is actually true as a statement of arithmetic.

Question 3

THE CALCULUS ANSWER (FOR THE BOLD)

One studies infinite sums in a calculus class. You learn there, for example, that 1+x+x2+x3+...=11x1+x+x{2}+x{3}+... = \dfrac {1} {1-x} is true as a statement of arithmetic for small values of xx, (specifically, for all values of xx strictly between 1-1 and 11.) The formula is valid for x=0.1x=0.1, as we saw, and not for x=2x=2.

For those who are game, here is how the calculus argument goes.

Regular polynomial division shows that 1x21x=1+x\dfrac {1-x^2} {1-x}=1+x and 1x31x=1+x+x2\dfrac {1-x^3} {1-x} = 1+x+x^{2} and 1x41x=1+x+x2+x3\dfrac {1-x^4} {1-x} = 1+x+x{2}+x{3}, and so on. (Try these!)
In general, we see that
1+x+x2+...+xn1=1xn1x1+x+x{2}+...+x{n-1}=\dfrac {1-x^n} {1-x}.

Now as we let nn get bigger and bigger it looks like we’re getting the infinite geometric sum.
1+x=1x21x1+x+x2=1x31x1+x+x2+x3=1x41x1+x+x2+x3+...=?\begin{aligned}
1+x &= \dfrac{1-x^2}{1-x}\
1+x+x^2 &= \dfrac{1-x^3}{1-x}\
1+x+x2+x3 &= \dfrac{1-x^4}{1-x}\
\vdots\
1+x+x2+x3+... &= ?
\end{aligned}

So the question is: What does 1xn1x\dfrac {1-x^n} {1-x} go to as nn gets bigger and bigger? If there is an answer to this question, then the answer will be the value of 1+x+x2+x3+...1+x+x{2}+x{3}+....

So does 1xn1x\dfrac {1-x^n} {1-x} have a limit value? Well, this depends on whether or not xnx^{n} has a limit value as nn grows. So for which values of xx do the powers of it approach a value?

We know the that powers of 0.10.1, for example, and of 0.830.83 and of 12- \dfrac {1}{2}, all approach zero for bigger and bigger powers. In fact, xnx^{n} gets closer and closer to zero as nn grows for any value between 1-1 and 11.

So for 1<x<1-1<x<1, we have
1+x=1x21x1+x+x2=1x31x1+x+x2+x3=1x41x1+x+x2+x3+...=101x=11x\begin{aligned}
1+x &= \dfrac{1-x^2}{1-x}\
1+x+x^2 &= \dfrac{1-x^3}{1-x}\
1+x+x2+x3 &= \dfrac{1-x^4}{1-x}\
\vdots\
1+x+x2+x3+... &= \dfrac{1-0}{1-x} = \dfrac{1}{1-x}
\end{aligned}

The geometric series formula can be believed, as a statement of arithmetic, for 1<x<1-1<x<1, at least.

Question 4

THE HONEST APPROACH

Another approach to examining the geometric series formula is to make an explicit leap of faith. Let’s assume that the infinite sum 1+x+x2+x3+...1+x+x{2}+x{3}+... is meaningful and has an answer. Call that answer SS.

Then

S=1+x+x2+x3+...=1+x(1+x+x2+...)=1+xS\begin{aligned}
S &= 1+x+x{2}+x{3}+...\
&= 1+x \left(1+x+x^{2}+...\right) \
&= 1+xS
\end{aligned}

from which we get S=11xS= \dfrac{1}{1-x}.

This is an honest approach. It proves:
IF the infinite sum 1+x+x2+...1+x+x^{2}+... has an answer, then that answer must be 11x\dfrac{1}{1-x}. It makes no assertion as to whether or not the infinite sum is meaningful in the first place.

The same is true for the dots-and-boxes approach as it too proves: IF 1+x+x2+x3+...1+x+x{2}+x{3}+... is meaningful to you, then it has the answer 11x\dfrac{1}{1-x}. It is up to you to decide if this infinite sum is meaningful.
(Calculus likes to say it is if 1<x<1-1<x<1.)

Question 5

OTHER SYSTEMS OF ARITHMETIC OFFER OTHER MEANINGS

The statement 1+2+4+8+...=11+2+4+8+...=-1 is meaningless in ordinary arithmetic. But who says we have to stay with ordinary arithmetic? Is there an extraordinary way to view matters?

We tend to view numbers as spaced apart on the number line additively. Walk one step to the right of 00 and we end up at position 11. Now add to that two steps and we end up position 33. Now add four steps, position 77. And so on. The sum 1+2+4+8+...1+2+4+8+..., in this viewpoint, takes infinitely far to the right of 00 on the number line.

1+2+4+8+...=1+2+4+8+...=\infty in ordinary arithmetic. It does not equal 1-1.

But let’s think of numbers multiplicatively. In particular, since we are focusing on the sum 1+2+4+8+...1+2+4+8+..., let’s think of factors and multiples of powers of two.

Now 00 is a highly divisible number. It is the most divisible number of all. With regard to just two-ness it can be divided by 22 once, in fact twice, in fact thrice. In fact, you can divide 00 by two as many times as you like—and still keep dividing.

With regard to two-ness, the number 88 is somewhat zero-like: you can divide it by two three times. But 3232 is even more zero-like: you can divide it by two five times. And 21002^{100} is even more zero-like still.

So in this sense, 21002^{100} is a number very close to 00. The number 3232 is somewhat close to 00. The number 88 is less close. The number 11 is not very close to zero at all: it cannot be evenly divided by two even once.

So, in this context, could 1+2+4+8+...1+2+4+8+... possibly be 1-1?

Well

1+2=3=411+2+4=7=811+2+4+8=15=161=1+2++299=21001\begin{aligned}
1+2 = 3 &= 4-1\
1+2+4 = 7 &= 8-1\
1+2+4+8 = 15 &= 16-1\
\vdots &= \vdots \
1+2+\ldots + 2^{99} &= 2^{100}-1
\end{aligned}

These finite sums grow to become “a number very close to zero, minus one.” In the limit, the infinite sum thus has value 01=10-1 = -1, just as our formal arguments said it would be.

So in this multiplicative view of arithmetic, 1+2+4+8+...1+2+4+8+... is a meaningful quantity, and it does indeed have value 1-1. The geometric series formula is meaningful and correct for x=2x = 2 in this context.

The point is that our dots-and-boxes work shows what the answers to many infinite sums have to be, IF the infinite sum has meaning to you. It is thus up to you to decide what arithmetic context you want to play in and whether or not the infinite sum you are examining is meaningful in that context. If it is, then dots and boxes tells you its answer!

Infinitia
Vision byPowered by
Contact GMP:
  • The Global Math Project on Twitter
  • The Global Math Project on Facebook
  • Contact The Global Math Project
Contact BM:
  • Buzzmath on Twitter
  • Buzzmath on Facebook
  • Visit Buzzmath's Website
  • Read Buzzmath's blog!