
The sum of all the positive integers is not -1/12 - lisper
http://blog.rongarret.info/2014/01/no-sum-of-all-positive-integers-is-not.html
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dmunoz
Ahhhh.... this is one of my biggest pet peeves. We're using the same words to
mean wildly different things, and then arguing past each other about meaning.
Every discussion thread I have seen on the matter has quickly brought up the
relevant actual semantics, but people still want to argue.

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...](http://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-
formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-
continuation/)

~~~
twotwotwo
The really interesting point, or at least what struck me when I first heard
about this, is that there are alternative, useful, counterintuitive notions of
ideas things that folks might think they have some intuitive grasp of. There's
more to math (and not only the parts you _know_ you don't know about) than is
dreamt of in one's philosophy.

~~~
Iv
Math is mostly about exploring the consequences of alternative definitions for
notions we take for granted. In this case, it is the equality notion that is
used in a different fashion

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Almaviva
> Step 1: Let S1 = 1 - 1 + 1 - 1 ...

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.

~~~
lisper
Actually, I didn't know that. And it surprises me, actually. Intuitively I
would have guessed that converging infinite sums would be well-behaved under
normal algebraic manipulations. But you learn something new every day :-)

Can you point me to a reference?

~~~
jfarmer
The basic idea is pretty simple. If Σa_i is conditionally convergent — that
is, Σa_i converges but Σ|a_i| does not — then there are an infinite number of
both positive and negative a_i. See [1] below for a hint on how to prove this;
it's straightforward.

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?

~~~
xyzzyz
> If Σa_i is conditionally convergent then there are an infinite number of
> both positive and negative a_i.

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).

~~~
jfarmer
I left out lots of other details, too, because it was meant to be an outline
for the original commenter, not a proof. In any case,

    
    
      \sum_{i=1}^{\infty} \frac{(-1)^i}{i^2}
    

is not conditionally convergent, so it is not a counterexample to my original
statement. Here I take _conditionally convergent_ to mean "the series
converges, but it does not converge absolutely." :)

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.

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ColinWright
Oh for the love of - does this _really_ need to be said?

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.

~~~
thaumasiotes
Well... I don't understand the mindset of "of _course_ it's false!" The first
question that comes to my mind is "can I use normal-seeming algebra to make 1
+ 2 + 3 + ... sum to a number other than -1/12"? If it's trivial to make it
come out to anything at all, we can chalk it up to apparently reasonable
things going wrong; if it seems to be mysteriously easier to add it all up to
-1/12, it's worth investigating why that might be.

I mean, we can apply the good old formula for the geometric series,

    
    
        1 + x + x^2 + x^3 + ... = 1/(1-x)
    

to x=2, and we learn that

    
    
        1 + 2 + 4 + 8 + 16 + ... = -1
    

And before rejecting that as obviously absurd, maybe we should take a minute
to reflect on how the computers we're using right now represent the quantity
-1?

~~~
jonsen
Ha, what a coincidence.

It won't work for decimal 10s complement numbers:

For x=10 we get

    
    
      1 + 10 + 100 + 1000 + 10000 + ... = -1/9
    

And -1 in a decimal computer is

    
    
      9 + 90 + 900 + 9000 + 90000 + …
    

so … OMG!

    
    
      9*(1 + 10 + 100 + 1000 + 10000 + ...) = 9*(-1/9) = -1

~~~
thaumasiotes
> Ha, what a coincidence.

> 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.

~~~
jonsen
_I don 't get why you say this_

Just a narrative of my own experience thinking this through :-)

I thought "OMG" would tell you that, sorry.

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werner34
I didn't watch the numberphile video and just read this.

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

~~~
14113
That's the point! Using the algebra from the article:

(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.

~~~
werner34
Alright, got it. Thanks!

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coldcode
The answer is clearly 42. Monkeying with infinity allows you to derive any
answer you like and I like 42.

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bren2013
I didn't feel like signing up for a Blogger account, but I still wanted to
drop this here:

> 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...](http://en.wikipedia.org/wiki/Riemann_sphere#Arithmetic_operations)

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.

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funemployed
For the love of math, this!

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csomar

      = (1 - 1 + 1 - 1 ...) +
         (0 + 1 - 1 + 1 - 1 ...)
    
      = (1 + 0) + (1 - 1) + (1 - 1) ....
    

I can't _get_ this part.

~~~
lisper
You're just adding the elements of the two series pair-wise (which "works"
because addition is commutative and associative).

Here are a few additional intermediate steps:

    
    
        =  (1 + -1 +  1 + -1 ...)
         + (0 +  1 + -1 +  1 ...)
        
        = (1 + 0) + (-1 + 1) + (1 + -1) + (-1 + 1) ...

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rtpg
This is a non-proof based only off of notational styles.

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

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russelluresti
Ugh. First, the title of this article is wrong because in the article itself
the author says it IS -1/12, just not the way the video proved it.

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.

~~~
jedi_stannis
No, the title is right. The sum is not -1/12, there is no sum because it
diverges. The Ramanujan summation is not the real summation, its a way to
assign a sum to divergent series which can be useful in some cases.

