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Divergent Infinite Series

Date: 05/30/2003 at 03:17:12
From: Hong Yew
Subject: Euler's Infinite Series

I bought a book _Praise for The Mathematical Universe_, by William
Dunham, with a chapter on Euler's infinite series. A proof he outlined
that I could not follow is this:

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

      1 = 1 and the "a" basically cancelled out.

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

      it factors out (1-a).

(3)   1/1-a = 1 + a + a^2 + a^3....

divides (1-a) on both sides. I'm confused.

1/1-a is clearly between -1 and 0 for any a > 1. 

However, the infinite series 1 + a + a^2 + a^3... is infinite for 
a > 1.

They should not equate.

Date: 05/31/2003 at 12:37:39
From: Doctor Rob
Subject: Re: Euler's Infinite Series

Thanks for writing to Ask Dr. Math, Hong Yew.

Operations with divergent infinite series often yield bizarre results.  
The series Euler started with diverges. To see why, consider the 
sequence of its partial sums. It goes like this:

   1, 1-a, 1, 1-a^2, 1, 1-a^3, ...

The odd-numbered partial sums are all 1; the even ones are 1-a^(n/2).  
The definition of the sum of any infinite series is the limit of its 
sequence of partial sums. If that limit does not exist, then the 
series is said to diverge. If the above sequence of partial sums has a 
limit, it must be 1 (look at the odd partial sums), but then one would 
have to have

   lim (1-a^[n/2]) = 1

(from the even partial sums), if and only if

   0 = lim a^(n/2),

if and only if -1 < a < 1.  For |a| >= 1, you have divergence. Then
all series operations such as those performed by Euler are invalid,
and may lead to nonsensical results.  This is what you found.

Feel free to write again if I can help further.

- Doctor Rob, The Math Forum 
Associated Topics:
High School Sequences, Series

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