The sum of all the positive integers is not -1/12
blog.rongarret.info
blog.rongarret.info
It's a shame that ColinWright's post, linked by him below, got nuked by the moderators because it contains the most relevant information on the matter directly from an accredited mathematics professor, Terry Tao [0]. This is what should gratifies one's intellectual curiosity, not a series of posts that argue about the boring parts.
[0] http://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin...
* Terence Tao got the Fields Medal in 2006 http://en.wikipedia.org/wiki/Terence_Tao
* ColinWright's post is now undead.
Because it's not so clearly stated in the article, step 1 is already flawed.
> the algebraic rules that apply to regular numbers do not apply to non-converging infinite sums
I'm sure the author knows this but they don't even apply to all converging infinite sums. (Only absolutely converging ones.) E.g. 1 - 1/2 + 1/3 - 1/4 ... converges to a value, but you can also re-arrange the terms to get any value you want.
Why is it strange that this sum is ln 2?
High school algebra tells us that
1 - x + x^2 - ... + (-1)^n x^n = 1 + (-x) + (-x)^2 + ... + (-x)^n = (1 - (-x)^{n+1})/(1-(-x)).
Thus, for the infinite sum we have:
1 - x + x^2 - x^3 + ... = lim_{n -> inf} (1 - x + x^2 - ... + (-1)^n x^n) = lim_{n -> inf} (1 - (-x)^{n+1})/(1-(-x)) = 1/(1+x).
But then if you integrate the infinite sum term by term, you'll get:
\int (1 - x + x^2 - x^3 + ...) = x - x^2/2 + x^3/3 - ...
On the other hand, if you integrate 1/(1+x), you'll get precisely log(1+x). Now, one would want to argue that:
log(1+x) = x - x^2/2 + x^3/3 - ...
and here this is actually true, but in general this may not be true that the integral of the sum of infinite series is equal to sum of the integrals of each term -- what you need here is the notion of uniform convergence, but fortunately 1 - x + x^2 - ... converges uniformly.
But, since Leonard Euler wouldn't really be pedantic about stuff like this, as long as the result is correct, neither should the occasional math hacker. Thus, we put x = 1 in both sides of the equality and we get 1 - 1/2 + 1/3 - ... = log(2).
Apart from integrating term by term, there's one more problem with above reasoning, and there are bonus points for people who notice that (hint: uggc://ra.jvxvcrqvn.bet/jvxv/Nory'f_gurberz )
Can you point me to a reference?
Given that, let's say we want to rearrange the a_i in Σa_i to converge to some arbitrary real number M. Take all the positive a_i in order and call that set A_+ and take all the negative a_i in order and call that set A_-. From [1] we know that both A_+ and A_- have an infinite number of elements.
You then construct a new series like so. Take the fewest elements from A_+ (in the order they appear in the original series) such that their sum is greater than M. Next, add to that sum the fewest number of elements from A_- (also in order) such that this new sum is smaller than M. Go back to taking elements from A_+ until you're just larger M and then back to taking elements from A_- until you're just under M. Repeat this ping-pong maneuver ad infinitum to construct a new series.
You can prove that this new series converges (conditionally) to M.
[1]: Here's a hint. What can we say about Σ|a_i| if we know that Σa_i converges but has only a finite number of negative terms?
That's true, but that's not enough for your proof -- e.g. \sum (-1)^n/n^2 also has infinitely many positive and negative terms, but converges absolutely.
What you need is that sum of all elements in at least one of A_+, A_- (thus also the sum of all elements in the second one) diverges (the order you take the sum in doesn't matter here, as you surely know).
\sum_{i=1}^{\infty} \frac{(-1)^i}{i^2}
What's more, if Σa_i is conditionally convergent then the sums of both A_+ and A_- diverge. You're right that one has to use this fact in the full proof.
"A series is said to be conditionally convergent iff it is convergent, the series of its positive terms diverges to positive infinity, and the series of its negative terms diverges to negative infinity. […] The Riemann series theorem states that, by a suitable rearrangement of terms, a conditionally convergent series may be made to converge to any desired value, or to diverge. The Riemann series theorem can be proved by first taking just enough positive terms to exceed the desired limit, then taking just enough negative terms to go below the desired limit, and iterating this procedure. Since the terms of the original series tend to zero, the rearranged series converges to the desired limit. A slight variation works to make the new series diverge to positive infinity or to negative infinity."
The correct method is to calculate the partial sum of the first N term: S_N=sum_{n=1}_N (-1)^n. When N is even the amount of 1 and -1 are equal and the partial sum is 0. When N is odd then there is an additional 1 and the partial sum is 1. So the partial sum has no limit and then 1 - 1 + 1 - 1 ... is not convergent.
Conditionally convergent series like 1-1/2+1/3-1/4+1/5-1/6+ ... =ln(2) are much better. (The problem is that 1+1/2+1/3+1/4+1/5+1/6+ ... = infinity.) You can operate with them almost freely, but you can’t rearrange the order of an infinite number of term.
The unconditionally convergent series like 1+1/4+1/9+1/16+...=pi^2/6 are almost like finite summations. You can do whatever sensible operation you want and the result will be correct.
"they [normal rules of algebra] don't even apply to all converging infinite sums"
I know that 1 - 1 + 1... is not convergent, but that's a non-sequitur relative to the claim that you made.
> The unconditionally convergent series like 1+1/4+1/9+1/16+...=pi^2/6 are almost like finite summations. You can do whatever sensible operation you want and the result will be correct.
Yes, that's what I thought.
http://www.amazon.com/Calculus-Vol-One-Variable-Introduction...
Unfortunately, I don't know any online references out of the top of my mind.
Really?
I mean, of course it's false! It's an instructive example of how apparently reasonable things go wrong, and why you sometimes really, really need to pay attention to the details.
Do we really need to be told that the sum of all the positive integers is a negative fraction? Of course we don't.
I despair sometimes, I really do. I need to go away and spend some time in my happy place.
I mean, we can apply the good old formula for the geometric series,
1 + x + x^2 + x^3 + ... = 1/(1-x)
1 + 2 + 4 + 8 + 16 + ... = -1
It won't work for decimal 10s complement numbers:
For x=10 we get
1 + 10 + 100 + 1000 + 10000 + ... = -1/9
9 + 90 + 900 + 9000 + 90000 + …
9*(1 + 10 + 100 + 1000 + 10000 + ...) = 9*(-1/9) = -1> It won't work for decimal 10s complement numbers:
I don't get why you say this -- you go on to demonstrate the opposite?
This is just the other-side-of-the-decimal-point inverse of the well known 0.9999999... = 1 equality.
One of my favorite proofs for 0.9999... being equal to 1 goes like so: imagine subtracting it from 1. You'll get a number which has a 0 at every decimal place: 0.0000000.... Obviously, the number with a 0 in every decimal place is 0 itself.
Similarly, if you have your integer represented decimally as a bunch of coefficients (0 <= c < 10) of powers of 10, and all of the coefficients are 9, it's fairly straightforward to see that adding 1 will get you a new number for which all of the coefficients are 0. Since adding 1 to the original number gave us 0, we can treat the original number as -1.
Just a narrative of my own experience thinking this through :-)
I thought "OMG" would tell you that, sorry.
This is proper geek fun with serious points underneath and a good smattering of why it's actually valid to consider these things. This isn't established people with proper reputations setting out to con the unsuspecting public. There is solid math going on here. See Terry Tao's recent blog post[0] about why we can and should play with these things.
And now it's late here and I'm going off-line for a while. As I said elsewhere, I need to detach for a bit.
[0] https://news.ycombinator.com/item?id=7078744
Added in edit: Just checking before signing off for the night and I see that the submission of Terry Tao's blog entry on this topic has been killed. By mods, flags, whatever, I don't know. here's the link[1] if you're interested, and this is me, signing off.
[1] http://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin...
Depends on what you mean by "they" and "it". The numberphile folks undoubtedly know that they are playing fast and loose with the rules. The readers of Slate almost certainly don't know it.
> There is solid math going on here.
No, there isn't. There is solid math going on in complex analysis where you have concepts like analytic extensions that produce unintuitive but useful and (more importantly) consistent results. But that is NOT what this video is about. This video is about using high school math to "prove" a result that is simply not true under the rules of high school math. The only thing separating this from outright crackpottery is that the result they derive happens to look like one that can be legitimately derived under the rules of complex analysis and analytic extensions. But that's a mighty thin reed. It doesn't change the fact that they present the result as if it were true under the rules of high school math, and under those rules it isn't true.
EDIT: Terry Tao's article is excellent, and I am appalled (but, sadly, not surprised) that your submission was killed.
Hm, my own submission seems to have fallen off the front page awfully quickly.
EDIT2: I have been corresponding with an HN mod who informed me that my submission triggered the voting ring detector. (It was a false positive, which ought to worry someone at YC.) Also, the original Terry Tao submission has now been unkilled. I encourage you to upvote it.
What is the Ramanujan summation of S_3 = 1+1+1+ ... ? (With a liberal interpretation of the Wikipedia article, I think it’s 0.)
Where is the fault in 0 = S_3-S_3 = S_3-shift(S_3)=1+0+0+0+...=1 ?
Is the shift a legal operation? When?
But the Wikipedia article “said” that it is 0 (because 1^0+2^0+3^0+...=1+1+1, and 0 is even). But the [dead?!] article submitted by ColinWright from Tao’s blog says it’s -1/2. (I prefer not to disagree with Tao, just fixed Wikipedia.)
The problem is more deep. I should read the complete version of Tao’s article.
You want to focus on the idea of analytic continuation. That's where the magic happens.
[1] http://books.google.cz/books?id=Of5G0r6DQiEC&pg=PA53&dq=grat...
Can someone explain to me why I am allowed to add a padding zero at the start without taking it out at the end?
By adding it, I am giving the sum an offset, and I can kind of understand why adding zero is not a big deal, but they reasoned:
Then S1 + S1 = (1 - 1 + 1 - 1 ...) + (1 - 1 + 1 - 1 ...)
= (1 - 1 + 1 - 1 ...) + (0 + 1 - 1 + 1 - 1 ...)
= (1 + 0) + (1 - 1) + (1 - 1) ....
= 1 + 0 + 0 ... = 1
But I might as well leave the leading Zero out and argue that:
Then S1 + S1 = (1 - 1 + 1 - 1 ...) + (1 - 1 + 1 - 1 ...)
= (1 - 1 + 1 - 1 ...) + (1 - 1 + 1 - 1 ...)
= (1 + 1) + ( -1 - 1) + (1 + 1) ....
= (2 - 2) + (2 - 2) + (2 - 2)... = 0
(1 - 1 + 1 - 1 ...) = 0 + (1 - 1 + 1 - 1 ...)
and therefore:
(1 - 1 + 1 - 1 ...) = (0 + 1 - 1 + 1 - 1 ...)
And as they're mathematically the same (again, in the article's algebra), why not replace one with the other?
The problem is that when you do, and you sum the two infinite series using the "zipping" method (as in the numberphile video) the two equations equate to different results.
Because there is no "end". These are infinite series.
> But I might as well leave the leading Zero out and argue that:
Yes, exactly, that's the whole point. You cannot manipulate non-convergent series in this way if you want consistent results.
> My friend, Ray Sidney, with a PhD in mathematics from MIT explained away this nonsense rather succinctly.
> infinity + 7 = infinity
> If you minus infinity from both sides, you can tell the world 7 = 0!
http://en.wikipedia.org/wiki/Riemann_sphere#Arithmetic_opera...
infinity - infinity (along with a few others) is very often left undefined for this very reason--not to mention it lets you divide by zero without breaking math.
= (1 - 1 + 1 - 1 ...) +
(0 + 1 - 1 + 1 - 1 ...)
= (1 + 0) + (1 - 1) + (1 - 1) ....
Here are a few additional intermediate steps:
= (1 + -1 + 1 + -1 ...)
+ (0 + 1 + -1 + 1 ...)
= (1 + 0) + (-1 + 1) + (1 + -1) + (-1 + 1) ...Let s be a series , with s_n = sum (1 to n, 1) = n if S3 exists (i.e. lim n->infinity s_n exists), then the series (s-s) (where (s-s)_n = s_n - s_n) converges , and it converges towards S3 - S3 (=0).
> = (1 + 1 + 1 + 1 ...) - (0 + 1 + 1 + 1 ...) the series whose sum he's describing here is not (s-s), but another one entirely : u where u_1 = s_1, u_n+1= s_n+1-s_n
These are different series, so it is entirely reasonable for them to converge at different values, 1 doesn't equal 0.
The "real" reason this result isn't what we think it is(apart from "infinite sums are different" argument, which is a non-answer):
>Step 1: Let S = 1 + 2 + 3 + 4 ... What you're doing here firstly, is saying that "I assume that this sum converges, let S be the value it converges to".
So you end up proving (if the other steps weren't also flawed) that S = -1/12, all you're saying is that if the sum exists, then it is -1/12.
The issue here depends on what sort of thing you're working on. The "classic" definition of a limit (convergence of a series) does not work here, because you can prove that for all n, the sum of n numbers will always be at least -1/12 away from -1/12 (on account of it being positive), so it can't converge to -1/12 (hence S not existing).
However, there are techniques for assigning limits to divergent sums. these summability methods will give the same result as the classic analysis for convergent series, but will also give values for some divergent series.
It is similar to analytic continuation( f(x) doesn't exist, but a limit exists in x both from the left and to the right (named y), so we sometimes say that f(x)=y ), in that it allows us to extend the resolution domain slightly. But the classic definition no longer works.
The one used here is zeta function regularization, which consists in the following:
you have a series a , and a function Za(s) = (1/a1^s)+(1/a2^s)+(1/a3^s)+(1/a4^s)+.....
Za is only defined for certain values of s depending on the series. But for a serie representing a convergent sum, we know that Za(-1)= a1+a2+a3+... exists. So this method will give the same limit for convergent sums as the classical method.
For a series with a diverging sum (a_n)=n, Za(-1) doesn't exist, but the limit in -1 exists from both ends (by analytic continuation), so we extend the domain of Za to -1 by the value of this "limit" : -1/12
The article mentions Ramanujan summation, but Zeta regularization is actually a much more useful tool. In
And, yes, I know in math that HOW you get the answer is somehow seen as being more important than the answer itself, but really? You consider this attention damaging? This is like the people who complain about Mythbusters because it's not "real science".
Look, anything that gets people more interested in math is great, especially when it's a video as harmless as this one (it's not as if this video could actually impact someone's life or well-being).
So drop this "I was the cool form of uncool before uncool became a thing" attitude and just be happy that people are interested in math for a change.
Sorry, can't do it. Mythbusters actually is real science, and anyone who thinks it isn't doesn't understand science. But this video is not real math, notwithstanding that the result it derives looks the same as one that can be derived using real math.
"You cannot obtain a negative number by adding positive numbers."
Try and disprove me, I dare you.