10
$\begingroup$

I've recently run into a proof that claims that $1 - 1 + 1 - 1 + 1 - 1 ... = \frac{1}{2}$ that proceeds as follows:

Let $S = 1 - 1 + 1 - 1 + 1 - 1 + ...$. Then $$S = 1 - (1 - 1 + 1 - 1 + 1 - 1 ...) = 1 - S$$ Therefore, $2S = 1$, so $S = \frac{1}{2}$. QED

I was under the impression that this sum doesn't converge, and therefore this step of the proof is invalid:

Let $S = 1 - 1 + 1 - 1 + 1 - 1 + ...$

That is, it's not even valid to suppose that the sum is equal to some (implicitly real or complex) number $S$, and therefore the reasoning that follows is meaningless because $S$ doesn't exist and therefore has any properties we want it to have.

Is the reasoning I've given above correct? That is, is it correct for me to claim that the proof fails because $S$ doesn't exist?

(I've heard that there are techniques for evaluating divergent integrals by using complex analysis and Taylor series, so perhaps there's another way to prove that the summation is $\frac{1}{2}$; I just wanted to see whether my reasoning is sufficient to explain why this particular proof is incorrect. If the proof actually is correct, then I stand corrected!)

Thanks!

$\endgroup$
  • 1
    $\begingroup$ Yes you have pinpointed the flaw. $\endgroup$ – Cheerful Parsnip Jan 12 '14 at 2:45
  • $\begingroup$ The argument changed the order of infinitely many terms. This is not allowed, unless you know the series converges absolutely. $\endgroup$ – Lost1 Jan 12 '14 at 2:45
  • $\begingroup$ @Lost1: Dear Lost1, Bracketing off a finite number of terms like this is fine for a convergent series, whether or not the convergence is only conditional. (It is infinite rearrangements which lead to possibly different limits, or divergence, in the conditionally convergent case.) Regards, $\endgroup$ – Matt E Jan 12 '14 at 2:57
  • $\begingroup$ as long as both of those s values are supposed to be the same, you can make 1/2 reduce to that. but consider "s_0 = 1 - s_1", which implies "s_0 + s_1 = 1" and "s_1 = 1 - s_0". It's mutual recursion. You can generate an infinite sum from 0 over (-1)^x that adds up to any value you like. try s_0 = 1/3. i don't know if uses that expect 1/2 to be the value also have the constraint that s_0==s_1, which would mean that the original result is also legitimate. $\endgroup$ – Rob Jul 28 '15 at 3:19
9
$\begingroup$

If you have some method of attaching a sum to a series (e.g. Cesaro summation, as discussed in Ayesha's answer, or Abel summation) which is linear (i.e. the limit of a linear combination of two series coincides with the same linear combination of the limits), and if $1- 1 + 1 - 1 + \cdots$ is summable with respect to this method, then this argument shows that the value of the sum will have to be $1/2$.

By itself, this argument won't tell you whether a given summation method applies to your series, though.

$\endgroup$
5
$\begingroup$

The series of course doesn't converge - that doesn't mean it doesn't have a sum. Indeed, note that the Cesaro means of the series tend to 1/2, and thus the Cesaro sum is 1/2.

Consider the partial sums of the series $s_n$. Then we say the (C, 1) sum of the series is the limit $$\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n}s_k $$ if this limit exists.

For your series here, we have the partial sums $1, 0, 1. . .$ and thus the Cesaro means are $1, \frac{1}{2}, \frac{2}{3}. . . $ a sequence tends to 1/2. Your proof is a heuristic demonstration of this rigorous fact.

$\endgroup$
  • $\begingroup$ And just checking here - saying that "the Cesaro sum is $\frac{1}{2}$" means "if you define the infinite sum of a series as the limit of the arithmetic mean of the first n partial sums as n goes to infinity, then under that definition the sum is $\frac{1}{2}$. However, that doesn't mean that the sum actually is $\frac{1}{2}$." Is that a correct understanding? $\endgroup$ – templatetypedef Jan 12 '14 at 2:50
  • $\begingroup$ I'm not really sure what you're asking here. If you define the sum of a series to be the (C, 1) sum, then yes, the sum of your series actually is 1/2. $\endgroup$ – Ayesha Jan 12 '14 at 2:52
  • 1
    $\begingroup$ Oh, okay, I think I see what you're saying. So does this really boil down to how you define the infinite summation? If you define it as "the limit of the partial summations," then since it diverges, the limit doesn't exist and the proof as written isn't correct. If you define it as the Cesaro sum, then the sum does exist, but the proof still isn't a rigorous demonstration of this fact and just gives an intuition? $\endgroup$ – templatetypedef Jan 12 '14 at 2:54
  • 1
    $\begingroup$ Yes, that would be the correct interpretation. To see why the proof is heuristic, observe that we can get very different results by similar procedures. If we insert parentheses in that series, we can obtain the sums 1 and 0. $\endgroup$ – Ayesha Jan 12 '14 at 2:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.