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 |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/