Ramble, Part 36: Closer and Closer
Oct. 28th, 2007 05:23 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
stoutfellowIn the last Ramble, I gave arguments for the values of the sums of three infinite series. The three arguments were identical in form, differing only in minor details, but one of the three seems obviously correct and the other two obviously wrong. In fact, all three arguments are flawed, and I'll discuss the flaw and the resolution under the cut.
The flaw in the arguments lay in the very first step. To set x equal to some expression is to make x another name for whatever the expression names; but what if the expression does not, in fact, name anything? Whatever follows is immediately vitiated.
What must be kept in mind is that it is impossible to literally add up infinitely many numbers. You can add two numbers, and thus any finite collection of numbers - just add two at a time. This cannot be extended to an infinite collection, though. To speak of the sum of an infinite series is to speak metaphorically; the best one can hope for is to achieve something which resembles addition well enough. The metaphor cannot be trusted, but must be tested. (For example, if you've taken calculus and remember the concept of conditional convergence, you may have seen that the ordinary commutative law cannot be generalized to infinite sums!)
The usual way of extending addition to infinite series involves the limit concept. In essence, it examines the successive finite sums - the sums of the first however-many of the series - to determine whether they approach some finite number. To say that a0+a1+a2+a3...=L is to say this: if you specify how close you want to get to L, there is some point past which all of the sums a0+a1+a2+...+an are that close to L.
To illustrate: in the case of the series represented by .333..., the successive sums are .3=3/10, .33=33/100, .333=333/1000, and so on. These do, indeed, approach 1/3; 3/10 differs from 1/3 by 1/30, 33/100 by 1/300, and so on. If you wish to get within one one-billionth of 1/3, you need only go to the ninth term 333,333,333/1,000,000,000 of the series, which differs from 1/3 by less than one one-billionth.
If we turn to the other two series, we see that the numbers 1, 1+2=3, 1+2+4=7, 1+2+4+8=15, and so on do not, in fact, approach any limit (much less -1); they grow without bound. The numbers 1, 1-1=0, 1-1+1=1, and so on do not grow without bound, but they don't approach any limit either - they alternate between 1 and 0.
Thus the reason why the first argument seems to work is that .333... does have a limit, while the other two do not. Once it is known that .333... does, in fact, name something, then the argument goes through: what it names must be 1/3. We can go further, then: if the series 1+2+4+8+... named anything, it would have to be -1, and if the series 1-1+1-1+... named anything, it would have to be 1/2.
Both of these are obviously wrong, are they not? Well... stay tuned.
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The flaw in the arguments lay in the very first step. To set x equal to some expression is to make x another name for whatever the expression names; but what if the expression does not, in fact, name anything? Whatever follows is immediately vitiated.
What must be kept in mind is that it is impossible to literally add up infinitely many numbers. You can add two numbers, and thus any finite collection of numbers - just add two at a time. This cannot be extended to an infinite collection, though. To speak of the sum of an infinite series is to speak metaphorically; the best one can hope for is to achieve something which resembles addition well enough. The metaphor cannot be trusted, but must be tested. (For example, if you've taken calculus and remember the concept of conditional convergence, you may have seen that the ordinary commutative law cannot be generalized to infinite sums!)
The usual way of extending addition to infinite series involves the limit concept. In essence, it examines the successive finite sums - the sums of the first however-many of the series - to determine whether they approach some finite number. To say that a0+a1+a2+a3...=L is to say this: if you specify how close you want to get to L, there is some point past which all of the sums a0+a1+a2+...+an are that close to L.
To illustrate: in the case of the series represented by .333..., the successive sums are .3=3/10, .33=33/100, .333=333/1000, and so on. These do, indeed, approach 1/3; 3/10 differs from 1/3 by 1/30, 33/100 by 1/300, and so on. If you wish to get within one one-billionth of 1/3, you need only go to the ninth term 333,333,333/1,000,000,000 of the series, which differs from 1/3 by less than one one-billionth.
If we turn to the other two series, we see that the numbers 1, 1+2=3, 1+2+4=7, 1+2+4+8=15, and so on do not, in fact, approach any limit (much less -1); they grow without bound. The numbers 1, 1-1=0, 1-1+1=1, and so on do not grow without bound, but they don't approach any limit either - they alternate between 1 and 0.
Thus the reason why the first argument seems to work is that .333... does have a limit, while the other two do not. Once it is known that .333... does, in fact, name something, then the argument goes through: what it names must be 1/3. We can go further, then: if the series 1+2+4+8+... named anything, it would have to be -1, and if the series 1-1+1-1+... named anything, it would have to be 1/2.
Both of these are obviously wrong, are they not? Well... stay tuned.
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