What we get if we add 1/2 infinite times

Question

I want to know if this is correct

We have this sums:

$$S1=1-1+1-1+1-1+1-1+1-1...=\frac12$$

$$S2=1-2+3-4+5-6+7-8...=\frac14$$

$$S3=1+2+3+4+5+6+7+8...=-\frac{1}{12}$$

If we take

$$S4=S1+S1+S1+S1+...=\frac12+\frac12+\frac12+\frac12...$$

we get

S4= +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 … +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 … +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 … +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 … +1 -1 +1 -1 +1 -1 +1 -1 … +1 -1 +1 -1 +1 -1 … +1 -1 +1 -1 … +1 -1 … + . . . +1 -1 +2 -2 +3 -3 +4 -4 +5 -5 +6 -6 +7 -7 +8 -8 …

we find that

$$S4=S3-S3=(1+2+3+4+5+6+7+8...)-(1+2+3+4+5+6+7+8...)=1-1+2-2+3-3+4-4+5-5+6-6+7-7+8-8...=\left(-\frac{1}{12}\right)-\left(-\frac{1}{12}\right)=-\frac{1}{12}+\frac{1}{12}=0$$

therefore

$$S4=S1+S1+S1+S1⋯=\frac12+\frac12+\frac12+\frac12...=0$$

And also, if we take:

$$S5=S2+S2+S2+S2...=\frac14+\frac14+\frac14+\frac14+...$$

we get

S5= +1 -2 +3 -4 +5 -6 +7 -8 +9 -10 +11 -12 +13 -14 +15 -16 … +1 -2 +3 -4 +5 -6 +7 -8 +9 -10 +11 -12 +13 -14 +15 … +1 -2 +3 -4 +5 -6 +7 -8 +9 -10 +11 -12 +13 -14 … +1 -2 +3 -4 +5 -6 +7 -8 +9 -10 +11 -12 +13 … +1 -2 +3 -4 +5 -6 +7 -8 +9 -10 +11 -12 … +1 -2 +3 -4 +5 -6 +7 -8 +9 -10 +11 … +1 -2 +3 -4 +5 -6 +7 -8 +9 -10 … +1 -2 +3 -4 +5 -6 +7 -8 +9 … +1 -2 +3 -4 +5 -6 +7 -8 … +1 -2 +3 -4 +5 -6 +7 … +1 -2 +3 -4 +5 -6 … +1 -2 +3 -4 +5 … +1 -2 +3 -4 … +1 -2 +3 … +1 -2 … +1 … + . . . = +1 -1 +2 -2 +3 -3 +4 -4 +5 -5 +6 -6 +7 -7 +8 -8 …

we find that

$$S5=S3-S3=(1+2+3+4+5+6+7+8...)-(1+2+3+4+5+6+7+8...)=1-1+2-2+3-3+4-4+5-5+6-6+7-7+8-8...=\left(-\frac{1}{12}\right)-\left(-\frac{1}{12}\right)=-\frac{1}{12}+\frac{1}{12}=0$$

therefore

$$S5=S2+S2+S2+S2...=\frac14+\frac14+\frac14+\frac14...=0$$

also

$$S5=S4$$

Can I sue the people who make videos on youtube saying to laymen that since $1 - 1 + 1 - \dots$ can be either +1 or 0, the value of the infinite sum is $1/2$? — Ant, Jun 28 '15 at 16:18
@Ant: how about saying that since $\lim\limits_{r\to-1^+}\sum\limits_{n=0}^\infty r^n=\frac12$ that $\sum\limits_{n=0}^\infty(-1)^n=\frac12$? — robjohn, Jun 28 '15 at 16:21
@robjohn Don't get your point.. Of course you can assign value to those expression (analytic continuation, cesaro sums etc) but that requires a solid understanding of what one is doing, and what "equality" means. But laymen don't have this kind of understanding and they get all confused when it turns out that contradictions start to appear as soon as you play with the series and their values ;) — Ant, Jun 28 '15 at 16:27
I upvoted this question because I don't want OP to be discouraged from asking questions and being curious about sequences and series in the future. It's not a bad attempt, and we want to encourage OP to deepen their understanding, not give up the subject. — A. Thomas Yerger, Jun 28 '15 at 16:32
@Ant: I was just asking how you felt about people who say that. That is harder to refute than simply saying it alternates between $0$ and $1$, so it is $\frac12$. I also dislike the flaccid (i.e. no rigor) popularization of regularization and analytic continuation that leads to these horrible statements. — robjohn, Jun 28 '15 at 16:33
@robjohn Well if they present it like an interesting obeservation / food for thought, I'd say that is an idea worth of more investigation. And upon doing so I believe one comes close to analytic continuation? It's a pity I never got the chance to study that properly, though :) If they present it as rock hard equality without any other comment, I'd be less happy :) — Ant, Jun 28 '15 at 16:40
I too support @Ant's position completely, as such videos are not making mathematics more accessible to laymen, but rather making it more like dreamland where you can make up your own rules however you like (when it is in fact diametrically opposite). Worse than not having those videos at all! And I applaud Serby for actually coming here to ask about these mathematical statements rather than blindly accepting. — user21820, Oct 11 '15 at 09:21
Serby, the important point is that with any claim there must be a crystal clear grasp of whether it has any real-world implications through some suitable interpretation. Anyone can invent infinitely many mathematical structures that have absolutely no meaning, just like anyone can invent infinitely many variants of chess. In the same manner one can invent a way of manipulating such infinite series, but one reaps what one sows. Sow nonsense and it is likely that one reaps contradictions. Even if no contradictions arise, there may be no meaningful interpretation of the results. — user21820, Oct 11 '15 at 09:25
In the latter case, one is only entitled to promote such manipulations as a game. I'm also aware of claims floating around that such divergent series appear in string theory and hence have a basis in the real world. Again it is important to be clear of what exactly the interpretation of series is. If it is the usual one, then certainly it is wrong. Instead it is often claimed that there are such things called cutoff functions that prevent divergence. However, these cutoff functions are chosen arbitrarily and fine-tuned to fit the experimental data, so draw what conclusions you want about them. — user21820, Oct 11 '15 at 09:30

score 3 · Answer 1 · answered Jun 28 '15 at 16:03

3

Every claim you started with is wrong. Your $S_{1}, S_{2}, S_{3}$ are all divergent series.

It looks like you declared these series to be equal to their Ramanujan summation assignment, (which is akin to setting a matrix equal to its own determinant), and then proceeded as if you actually meant it converged to that value.

answered Jun 28 '15 at 16:03

Jonathan Hebert

5,272

That's exactly what I did. I wanted to get as much info about this video as I could and I thought that this way of asking will get a lot of answers. By the way, I am not a math guy. – Serby Jun 28 '15 at 16:20
1

@Serby: those videos exceed the limits of rigor, and great care needs to be exercised when using the "results" gotten there. Without knowing exactly what is going on, those videos can be very misleading. – robjohn Jun 28 '15 at 16:27

score 1 · Answer 2 · answered Jun 28 '15 at 15:47

1

You start with wrong conditions.

For example, the infinite sum $\sum_n^{\infty} (-1)^n $ is not equal to $1/2$.

answered Jun 28 '15 at 15:47

Christian

1,781

score 1 · Answer 3 · answered Jun 28 '15 at 16:17

As the other answers have said, your reasoning is very off: these are divergent sequences, and don't behave the way you think they do.

On the other hand, a reasonable question to ask is: "Can we make sense of these divergent series?" The answer is yes, although it is not easy: there are several methods for summing (some) classically divergent series. See https://en.wikipedia.org/wiki/Cesàro_summation (according to which which $1-1+1-1 . . .={1\over 2}$, https://en.wikipedia.org/wiki/Borel_summation, or https://en.wikipedia.org/wiki/Divergent_series#Abel_summation.

As for the rest of your reasoning, you rely on properties of series which do not hold for divergent series, even when interpreted according to Cesaro or other - e.g., being able to freely rearrange and combine terms. So to be frank, your reasoning is far off. But there are very interesting things you can do with divergent series!

(An example of the bizarre properties of divergent series: according to Cesaro summation, $-1+1-1+1. . .=-{1\over 2}$, even though it's "clearly" the same as $1-1+1-1.. .$.)

score 1 · Answer 4 · answered Jun 28 '15 at 16:20

I think it is praiseworthy that you are interested in these infinite sums and are thinking of ways to combine them in order derive new results. However there is a problem: all these sums are not strictly convergent, but exist only if one extends a correct result to a case where they are not really properly defined. Unfortunately this means that the operations that you carry out are also dubious.

I suggest that you start with strictly converging sums. For example your $S4$ is best written as:

$$S4(t) = 1 - t + t^2 - t^3 + t^4 - t^5 + t^6-t^7+t^8-t^9+t^{10}........$$

where t is a positive real number smaller than 1. You can now consider what happens if you calculate $S4(t) + t^2S4(t) + t^4S4(t) + ....$ as in your example.

score -1 · Answer 5 · answered Jun 28 '15 at 16:18

-1

There is no official way to add these divergent alternating series, but $\sum (-1)^n$ is known as Grandi's series:

$$ \sum_{n=0}^\infty (-1)^n = \begin{cases} 1 & \text{ if }n\text{ is odd} \\ 0 & \text{ if }n\text{ is even} \\ \end{cases} \approx \frac{1}{2}$$

And for your second series $(1 - 1 + 1 - 1 + \dots)^2 = 1 - 2 + 3 - 4 + \dots = \frac{1}{4}$ Wikipedia gives this nice figure

Doesn't it seem obvious that $S5 = \frac{1}{2} S4$ ? However wikipedia gives the value $\frac{1}{2}$.

answered Jun 28 '15 at 16:18

cactus314

24,438

3

I down voted because the OP is clearly confused about what "equality" even means when it comes to infinite sum. With this answer you're actually reaffirming it's understanding that $1 - 1 + 1 - ... = 1/2$ – Ant Jun 28 '15 at 16:22
@Ant his question seems to be whether these divergent series identities can be found via elementary rearrangements. – cactus314 Jun 28 '15 at 16:35
1

Yes but I doubt he has a clear understanding of what a "divergent series identity" is.. If you ask me this will just confuse him further. I don't mean to discourage from the study of math but I don't like this "sensational" statements without any sort of understanding of what is going on. It makes math appear obscure, patently absurd, and arbitrary – Ant Jun 28 '15 at 16:38
1

@Ant There is nothing wrong with this kind of experimentation. Euler did it, Physicists do it routinely (via their lack of rigour). He just needs to understand these series come with at the expense of associativity. We can't invoke $(a+b)+c = a + (b+c)$ infinitely many times and always get away with it. Watch his parentheses carefully. – cactus314 Jun 28 '15 at 16:49
1

Well there are a lot of people who flight planes, too, but I wouldn't trust myself and a manual of instruction. I really feel like the only answer at this level for the OP is "the sum is divergent and it's meaningless to attach any value to it. Then come talk to me in a couple of years about analytic continuation " :-) Of course other may disagree! :) – Ant Jun 28 '15 at 17:08
@Ant Maybe the question sounds naive but the answer "the sum is divergent and it's meaningless to attach any value to it. Then come talk to me in a couple of years about analytic continuation " has no point. It is obvious that I posted the question as an "naive" form of experimentation. I was well aware of the meaning of divergent series identity when I posted the question. Since I don't have enough math knowledge I wanted to see what would be the effect of such experimentation from a broader mathematical point of view. – Serby Jun 28 '15 at 19:07
@Serby I am sorry if I've offended you, as it was not my intention. I also like when people care about math, I don't want to scare you off! Also, I still haven't studied analytic continuation well, so those series to me still are largely just divergent series. On the other hand, if you're aware of what divergent means and what analytic continuation is, it certainly does not appear so from you question, as you would know that some properties do not hold. If you want my advice, you can start studying it more rigorously to get to the bottom of the matter. – Ant Jun 28 '15 at 19:14
If you enjoy "naive" speculations (I do too!) there are plenty of problems less "controversial", so to speak, to have fun with.. ie contest math, number theory and the like. – Ant Jun 28 '15 at 19:14
You didn't offend me in any way. – Serby Jun 28 '15 at 19:32
Regarding those speculations, I still consider that my question was good enough for my purpose. It may look plain stupid but I wanted some, let's say, "unusual" but mathematical rigorous point of view about those infinite divergent sums. My curiosity was based on the way those sums are used in the String Theory and the fact that even nowadays mathematicians debate about the meaning of attaching a value to those sums.I got many more stupid results from experimentation with this sums in this "naive" manner but I thought that it wouldn't be appropriate to post those. – Serby Jun 28 '15 at 19:49
The cardinality of the continuum was always a concept that astonished me and invited me to play with the concept of infinity. These sums are striking the same string so I am willing to speculate as much as I can without any remorse. In other contexts (in math or other science) I tend to be a very rigorous person. – Serby Jun 28 '15 at 20:06

What we get if we add 1/2 infinite times

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