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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 + x + x^2 + x^3........

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 + 2 + 4 + 8 + 16.......

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

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

Thanks for any help you provide.

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 

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 
understand your answer.

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,

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

Hi Peter,

An infinite series


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 

       |(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


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 

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


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 
Associated Topics:
High School Sequences, Series

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