1 + 2 + 3 + 4 + .... = -1/12 ? (infinite series)

[excreationist]

For a while I've heard that the sum of 1 + 2 + 3 + 4 +.... = -1/12

e.g. (with 8 million views)

The answer to this sum is -- remarkably -- minus a twelfth.​

It's amazing! I mean, I first saw this result when I start learning a bit of String Theory​

And what's even more bizarre is that this result is used in many areas of physics​

This is a very well known string theory textbook by Joe Polchinski. As you can see,​

sort of quite early on, page 22, we have this statement here which is that the sum of​

all this is --- basically saying the sum of all the integers ---​

all natural numbers all the way up to infinity, is, minus a twelfth.
​

Well I've never even suspected that the sum could be -1/12...

It looks like my hunch could be correct - the sum seems to be in fact infinity...



That video is very unhappy with the Numberphile math but ends up using a Riemann zeta function to show that 1 + 2 + 3 + 4 +... is connected with -1/12...

Though 25 minutes in it says the Riemann zeta function "makes sense if the real part of z is greater than 1". You can get 1 + 2 + 3 + 4.... connected with -1/12 if you substitute z with -1. But "we really don't have equality here".

This agrees with what Numberphile said about string theory:

https://en.wikipedia.org/wiki/Riemann_zeta_function

zeta(-1) = -1/12
This gives a pretext for assigning a finite value to the divergent series 1 + 2 + 3 + 4 + ⋯, which has been used in certain contexts (Ramanujan summation) such as string theory.​

 

[Swammerdami]

Mathologer is much better than Numberphile, though it often takes a lot more work to follow.

It is quite "neat" that knowledge of any interval of an analytic function, no matter how tiny, uniquely implies the entire analytic function.

 

[steve_bank]

An ifin9te series by definition can not be expressed as a finite value.

Someone on the forum argued that 0.9999.... = 1 based on the misapplication of a technique.

 

[excreationist]

An ifin9te series by definition can not be expressed as a finite value.

Someone on the forum argued that 0.9999.... = 1 based on the misapplication of a technique.

I would say 0.9999.... approaches 1...

 

[Artemus]

Here we go again. For those who really want to understand, start at  0.999.... Note the proofs going all the way back to Euler. If you still believe that you can show that 0.999... != 1, submit your proof to an appropriate journal and wait for your Nobel prize in Mathematics. Let us know how the peer review turns out.

 

[Jimmy Higgins]

Here we go again. For those who really want to understand, start at  0.999.... Note the proofs going all the way back to Euler. If you still believe that you can show that 0.999... != 1, submit your proof to an appropriate journal and wait for your Nobel prize in Mathematics. Let us know how the peer review turns out.

Aren't there things like 0.99... = 1.0, but not that specifically, that appear to show contradictions or two unequal numbers equaling each other indicating the math indeed was done wrong? Yes, 0.999... = 1.

I saw one video on YouTube showing something like that, and the class was like wow, and during the process I said, 'you can't do what he just did'... and then when done, he showed what was done incorrectly, going back to where I thought there was an error, sorry I can't remember what it was.

 

[Artemus]

Here we go again. For those who really want to understand, start at  0.999.... Note the proofs going all the way back to Euler. If you still believe that you can show that 0.999... != 1, submit your proof to an appropriate journal and wait for your Nobel prize in Mathematics. Let us know how the peer review turns out.

Aren't there things like 0.99... = 1.0, but not that specifically, that appear to show contradictions or two unequal numbers equaling each other indicating the math indeed was done wrong? Yes, 0.999... = 1.

I saw one video on YouTube showing something like that, and the class was like wow, and during the process I said, 'you can't do what he just did'... and then when done, he showed what was done incorrectly, going back to where I thought there was an error, sorry I can't remember what it was.

A classic algebraic "proof" that 1 = 2 I first saw in junior high school algebra is given at  Division_by_zero#Fallacies. The division by x-1 is invalid since x-1=0.

 

[Swammerdami]

I'd seen that Mathologer video before, but rewatched it and saw something I'd missed the first time. The Ramanujan summation 1+2+3+4+... = -1/12 was known long before Ramanujan. I thought Ramanujan just rediscovered it, but in fact the details of Ramanujan's clever summation method went beyond what was previously known.

Before writing his famous letter to Hardy, Ramanujan had written to a different English mathematician. That guy assumed 1+2+3+4+... = -1/12 was foolishness and wrote a condescending letter back. It's a good thing Ramanujan didn't give up, and had the courage to write Hardy, the top English mathematician of that time.

~~~~~~~~~~~

On the topic of 1 = 0.999999... I hope we don't have any TFT'ers that dispute that! One idea that might help one's intuition is to note that 1/4 has TWO decimal renderings (0.250000.... and 0.249999...) while 1/3 has only one (0.333333...) If there really is a meaningful difference between the two renditions of 1/4, why does 1/3 have only one? And what about base-9 where the vice versa applies? (Then it is 1/3 and not 1/4 with two renderings.)

The pretentiously-titled Axiom of Archimedes is another way to know that 1 and 0.99999.... must be equal. If not, their difference would be non-zero but would remain infinitesimal after multiplication by any integer, however large. This violates the Axiom of Archimedes. That Axiom, actually postulated before Archimedes, was used to dispose of Zeno's paradoxes (and for other purposes).

Some mathematicians have amused themselves by designing "Non-Archimedean Arithmetic (Fields)". But 1 ≠ 0.99999.... — that ain't it.

 

[lpetrich]

The  Riemann zeta function is defined as

\( \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^{s-1}}{e^x - 1} dx \)

where Γ(s) is given by

\( \Gamma(s) = \int_0^\infty x^{s-1} e^{-x} dx \)

and Γ = (n-1)! for n a positive integer.

With this definition, ζ(s) is defined only for Re(s) > 1. But near 1, it has the value

\( \zeta(s) = \frac{1}{s-1} + \gamma + O(s-1) \)

where γ is the Euler-Mascheroni constant, 0.577216...

This result suggests that ζ(s) could be defined for s < 1, but doing so will require going around s = 1 in the complex plane -- analytic continuation.

For a hint at that, consider a similar function:

\( \eta(s) = \sum_{n=1}^\infty \frac{(-1)^{s+1}}{n^s} = \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^{s-1}}{e^x + 1} dx = (1 - 2^{1-s}) \zeta(s) \)

The nice thing about η(s) is that it is defined for Re(s) > 0, and one can thus gets ζ(s) for Re(s) > 0.

One can get ζ(s) for s < 0 with Riemann's reflection equation

\( \zeta(s) = 2^s \pi^{s-1} \sin (\pi s /2) \Gamma(1-s) \zeta(1-s) \)

That makes ζ(s) defined over the entire complex plane.

The gamma function also has a reflection equation

\( \Gamma(s) \Gamma(1-s) = \frac{\pi}{\sin (\pi s)} \)

So that's how one gets ζ(-1) = - 1/12.

 

[steve_bank]

The technique used to prove .999... = 1 is intended to yield an approximation for a decimal. t does not equate 0.99..., to be numerically equal to 1. When you plug 0.999... into the alogithm the only possible result is 1.

Plug in 0.2222 or 0.31675 and see what happens.


When confronted with problems I can not get a handle on I resort to fundamentals.

First 0.999.. is not a number by definition. From high school math the equal sign implies a blame, like a balance scale. Each side nus be the same numerically.

On the face of it equating 0.999... to an integer 1 has no meaning.

Like all logic one can prove any number of assertions in a consistent proof, but is not realizable in reality. A valid syllogism alone does not make the argument valid.

I can write a valid differential equation for an electrical circuit that violates LOT, and actually simulate it. Mathematically the circuit runs forever without running down. Does that disprove entropy?

In the end math is as empirical as physical science. Many mathematically consistent theories have come and gone proven invalid in reality.

Show me where 1,2, 3... in a real world example where the substitution applies. If not I give it a 10 for artistic creativity and elegance, 0 for anything else.

To be honest the op is laughable.

 

[steve_bank]

An ifin9te series by definition can not be expressed as a finite value.

Someone on the forum argued that 0.9999.... = 1 based on the misapplication of a technique.

I would say 0.9999.... approaches 1...

Not really. If a result in a series of calculations is 0.999... you can choose to round up depending on the situation. Ther4 are situations where you would not want to round up.

The limit of 1/(1 + (1/x)) as x goes to infinity is 1. It does not equate the series to 1 but says under certain conditions you can simplify by subsuming 1 in calculations.A useful technique in many kinds of problems.

 

[Swammerdami]

First 0.999.. is not a number by definition. From high school math the equal sign implies a blame, like a balance scale. Each side nus be the same numerically.

Wrong. 0.99999999... is an alias for 1.00000..., two ways to write the same number.
"0.999.. is not a number" ?? - Why not? Because the digits go on forever? pi = 3.14159265... has digits that go on forever. Is it also not a number?

As the "academic credential" for my claims, note that Brady had me consult for one of his more popular Numberphile episodes.

 

[steve_bank]

I looked at the algorithm. One use is to generate fractional approximations to decimals.

You can do the calculations by hand. Plug in .1111... , .2222... , …. .99999.. and see what you get.

The fractional approximation to o.9999.. from the algorithm is actually 1/1 which of courses is 1. It does not say 0.999.. and 1 are exactly interchangeable. The fractional equivalent of 0.333... is 1/3. Look at what what the an algorithm is supposed to do.

As to the op when in doubt run test cases aghast your math.

1/[1+2+3....] ---> 0 in the limit

1/[-1/12] = -12

0.9999... does not equal -1/12. It may have some meaning in a larger theoretical context, but computationally it is not equivalent.

Regardless of any math all you have to do is show one counter example.

As to there ne no meaningful difference between 0.9999.. and 1 it depends on the series of calculations, accuracy, and error propagation.

 

[excreationist]

I would say 0.9999.... approaches 1...

Not really. If a result in a series of calculations is 0.999... you can choose to round up depending on the situation. There are situations where you would not want to round up.

It's 0.9 then 0.99 then 0.999 then 0.9999.... it's getting closer and closer to 1.... it is approaching 1 more and more....

 

[Artemus]

I would say 0.9999.... approaches 1...

Not really. If a result in a series of calculations is 0.999... you can choose to round up depending on the situation. There are situations where you would not want to round up.

It's 0.9 then 0.99 then 0.999 then 0.9999.... it's getting closer and closer to 1.... it is approaching 1 more and more....

That's not a correct interpretation. The 9s are not added one by one, they are simply there. See the Wiki article I linked on the first page. It covers the objections that have been raised.

 

[bilby]

1/3 = 0.333...

0.333... x 3 = 0.999... = 1

If 0.999... != 1, then 1/3 x 3 !=1, which is a contradiction.

If 0.999... <1, you could repeatedly divide by 3 then multiply by 3, and get a smaller and smaller result with each iteration.

Being proud of having the "common sense" to be incapable of grasping this simple proof is truly sad.

 

[Artemus]

1/3 = 0.333...

0.333... x 3 = 0.999... = 1

If 0.999... != 1, then 1/3 x 3 !=1, which is a contradiction.

If 0.999... <1, you could repeatedly divide by 3 then multiply by 3, and get a smaller and smaller result with each iteration.

Being proud of having the "common sense" to be incapable of grasping this simple proof is truly sad.

The (incorrect) objection that will be raised shortly is that 0.333... does not equal 1/3 but only approaches it. For some reason people have difficulty accepting that a convergent infinite series has an exact value, even though that was proven centuries ago and is in fact why Achilles does indeed catch the tortoise.

 

[bilby]

1/3 = 0.333...

0.333... x 3 = 0.999... = 1

If 0.999... != 1, then 1/3 x 3 !=1, which is a contradiction.

If 0.999... <1, you could repeatedly divide by 3 then multiply by 3, and get a smaller and smaller result with each iteration.

Being proud of having the "common sense" to be incapable of grasping this simple proof is truly sad.

The (incorrect) objection that will be raised shortly is that 0.333... does not equal 1/3 but only approaches it. For some reason people have difficulty accepting that a convergent infinite series has an exact value, even though that was proven centuries ago and is in fact why Achilles does indeed catch the tortoise.

Well at least such an objector can console themselves with the fact that it's impossible for them to fall out of a tree.

 

[Jarhyn]

1/3 = 0.333...

0.333... x 3 = 0.999... = 1

If 0.999... != 1, then 1/3 x 3 !=1, which is a contradiction.

If 0.999... <1, you could repeatedly divide by 3 then multiply by 3, and get a smaller and smaller result with each iteration.

Being proud of having the "common sense" to be incapable of grasping this simple proof is truly sad.

The (incorrect) objection that will be raised shortly is that 0.333... does not equal 1/3 but only approaches it. For some reason people have difficulty accepting that a convergent infinite series has an exact value, even though that was proven centuries ago and is in fact why Achilles does indeed catch the tortoise.

Or that any value can be expressed as the convergence of an infinite series. Isn't that one of the basic axioms of math, the Law of Identity?

 

[steve_bank]

The practical problem is that all math operations are finite and truncated, either digital or by hand calculation. The order of calculation on a calculator can affect raccuracy.

Before calculators using fractions instead of decimals was a way to -reserve accuracy and make calculations easier..

After thinking abo9ut it last night my problem with the op assertion is the use of =.

y = x +1
f(1) = 2 means that the operation on 1 results in 2 , it does not mean 1 = 2.

For the op f(1+2+3..) = -1/12. 1+2+3... does n0t equal -1/12