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### Sum of a Geometric Progression

```Date: 12/01/2002 at 07:55:22
From: Peter
Subject: Infinite sum

Dear Doctors,

I found this strange equation on the internet, but didn't find any

1
----- = 1 + x + x^2 + x^3........
(1-x)

It's not hard to verify that:

(1-x)(1 + x + x^2 + x^3........) = 1 - x + x - x^2 + x^2..... = 1

Then this equation must be true for any number x <> 1

Let's try to plug the number 2 instead of x:

1
----- = -1 = 1 + 2 + 4 + 8 + 16.......
(1-2)

It's really strange that an infinite sum of positive numbers is
negative!

Please could you help me in solving the problem or finding the error
in the equation?

Sincerely,
Peter
```

```
Date: 12/01/2002 at 08:37:06
From: Doctor Jerry
Subject: Re: Infinite sum

Hi Peter,

The series

(1-x)^{-1} = 1+x+x^2+...

converges only if |x| < 1.  Otherwise, it does not.

The series can be seen as the sum of a geometric progression.  We
know that

a+a*r+a*r^2+... = a/(1-r), provided that |r| < 1.  Here, a=1 and r=x.

- Doctor Jerry, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 12/01/2002 at 08:43:40
From: Peter
Subject: Infinite sum

Dear Dr. Jerry,

Thanks for responding to my question, but unfortunately, I didn't

1) why does (1-x)^(-1)=1+x+x^2.... converge only if |x|<1 ?
2) why does a+a*r+a*r^2+... = a/(1-r) provided that |r|<1 ?

Thanks for any help,
Peter
```

```
Date: 12/01/2002 at 15:12:18
From: Doctor Jerry
Subject: Re: Infinite sum

Hi Peter,

An infinite series

a1+a2+a3+...

of numbers is said to converge provided that there is a number S such
that for every positive number E there is a positive integer N for
which

|(a1+a2+...+an) - S | < E

whenever n >= N.  The number S is called the sum of the series.  This
definition is a standard definition of convergence.  You can find it
in most calculus books.

For the series

1+x+x^2+...

it is clear that it does not converge to a number S when x = 1. The
sums 1+x+...+x^{n-1} = n and hence grow beyond all bounds as n
increases.

We can form the partial sum 1+x+...+x^{n-1} of the first n terms as
follows: denoting this partial sum by Sn,

Sn = 1+x+...+x^{n-1}

x*Sn = x+x^2+...+x^n

Sn - x*Sn = 1 - x^n

Hence,

Sn = (1-x^n)/(1-x).

If |x|<1, we see that the limit of the numerator is 1.  Hence,

Sn approaches 1/(1-x)  as n increases without bound.

In the definition of convergence set out above, we take S=1/(1-x); in
this case,

|S-(1+x+x^2+...+x^{n-1}| = |1/(1-x) - (1-x^n)/(1-x)| = |x^n/(1-x)|

With x fixed and |x| < 1, if E is any positive number, all we have to
do is to take N sufficiently large that |x^N| < E(1-x).  The rest
follows easily.

- Doctor Jerry, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Sequences, Series

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