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A Series that Converges and Diverges?

Date: 04/30/2002 at 03:49:53
From: Darren Stanley
Subject: A series that converges AND diverges?

I'm told that this question is related to a new branch of 
mathematics. Thus, I cannot even begin to describe what sort of 
question my question is. Nevertheless, it involves some elementary 
mathematics. Here it is:

Let N = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + ...

The series diverges, right?

But N = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + ...
          = 1 + 2(1 + 2 + 4 + 8  16 + 32 + 64 + ...)
          = 1 + 2(N)

So...if N = 1 + 2N, then N= -1

So, then the series converges...

Huh?
Thanks.


Date: 04/30/2002 at 08:36:14
From: Doctor Pete
Subject: Re: A series that converges AND diverges?

Hi,

Your question is an interesting one; however, I am not aware of it 
being related to anything particularly novel. The infinite series you 
describe as N is in fact divergent, and the argument you provide 
showing N is finite is false.

To see why, we first appeal to the definition of an infinite series.  
For all nonnegative integers n, consider the sequence

     {a[n]} = {a[0], a[1], a[2], a[3], ... }.

Then the infinite series which we usually write as

     S = a[0] + a[1] + a[2] + a[3] + ...

is defined to be

     lim S[k]

as k -> infinity, where is the finite partial sum

     S[k] = a[0] + a[1] + ... + a[k].

It is in this manner that the precise meaning of an infinite series 
is made clear.  Thus, when we look at

     N = 1 + 2 + 4 + 8 + ...

we are really looking at the limit of partial sums of the sequence

     {a[n]}, n >= 0,  a[n] = 2^n.

It is quite obvious that the sequence is unbounded above, and 
therefore the limit fails to exist and N is what we call "divergent."

On the other hand, we need to understand why the "proof" of the 
convergence of N is erroneous.  The mistake is a subtle one:  The 
relation

     N = 1 + 2 + 4 + ... = 1 + 2(1 + 2 + 4 + ...)
       = 1 + 2N

from which we obtain the equation

     N = 1 + 2N,

is, in the most general sense, not the problem. The problem occurs 
when you subtract N from both sides and proceed to manipulate the 
equation algebraically to conclude N = -1. These operations cannot 
be done because they presume the value of N to be well-defined (in 
this case, finite). In essence, the conclusion N = -1 is based on the 
assumption that N converges; and this is not allowed, because one 
cannot assume what is not yet proven (and is in fact false). Hence 
the false assumption leads to a meaningless result.

For another example of some incorrect reasoning about infinite 
series, consider the series defined by the sequence

     a[n] = (-1)^n,  n >= 0.

That is, let

     M = 1 - 1 + 1 - 1 + 1 - 1 + ... .

If you pair successive terms in the following manner,

     M = (1 - 1) + (1 - 1) + (1 - 1) + ... = 0.

But then, if we pair the terms differently,

     M = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ... = 1.

This seemingly paradoxical result is simply the result of neglecting 
the definition of an infinite series. We are not allowed to compute 
the infinite sum M by selectively grouping its terms, but instead we 
must write

     M = lim S[k]

where k -> infinity, and

     S[k] = (-1)^0 + (-1)^1 + ... + (-1)^k,

from which we find that S[k] = 0 if k is odd, and 1 if k is even. In 
other words, we have

     S[k] = 1 + (-1)^k

for all k, and hence

     M = lim S[k] = 1 + lim (-1)^k,

but this last limit is undefined (it may seem obvious, but a rigorous 
proof is a topic for real analysis and the definition of a limit).  
Therefore, M is divergent. But it is important to note that S[k] is 
bounded; thus M is not divergent in the sense that its value is 
infinite, but rather because it has no well-defined finite value. 
That is, we always have

     0 <= S[k] <= 1,

but lim S[k] is not a fixed number in this interval.

I hope this discussion has cleared up some of your questions.  To 
have a full appreciation of the topics we've brought up, research the 
subject of real analysis, and in particular, the idea of sequences, 
series, and limits.

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


Date: 04/30/2002 at 13:26:47
From: Darren Stanley
Subject: A series that converges AND diverges?

Thank you for responding to my question on N = 1+2+4+8+16+... 
converging/diverging. You may be amused that this very question 
stumped a large roomful of mathematicians and math educators at a 
PIMS conference last week in Vancouver, BC.

Thank you once again.
Darren
Associated Topics:
High School Sequences, Series

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