Associated Topics || Dr. Math Home || Search Dr. Math

### 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

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search