Intuition for 1 + 2 + 3 + … = -1/12

In precisely the same sense that we can write 1 + 1/2 + 1/4 + ... = 2, despite that no real-world process of "addition" involving infinitely many terms may be performed in a finite number of steps, we can write 1 + 2 + 3 + 4 + 5 + ... = -1/12

I think this is overstating things (which is fair enough to make the point you're making).

The first is simply a shorthand for "the limit of this sum is 2", which is an extremely simple, general definition, which applies in almost all contexts, and matches up with what addition means in almost all contexts. It preserves far more of the properties of addition as well - it's commutative, associative, etc. In most cases where you want to work with the sum of an infinite series, the correct value to use for this series is 2.

The second is a shorthand for something far more complex, which applies in a far more limited range of cases, and doesn't preserve almost any of the properties we expect of addition. It's not linear or stable. In most cases where you want to work with sums of infinite series, the correct sum for this series is infinity. Only very rarely would you want -1/12.

 In precisely the same sense that we can write

$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\dots =2$,

despite that no real-world process of "addition" involving infinitely many terms may be performed in a finite number of steps, we can write

$1+2+3+4+\dots =-\frac{1}{12}$.

Well, not precisely. Because the first series converges, there's a whole bunch more we can practically do with the equivalence-assignment in the first series, like using it as an approximation for the sum of any finite number of terms. -1/12 is a terrible approximation for any of the partial sums of the second series.

IMO the use of "=" is actually an abuse of notation by mathematicians in both cases above, but at least an intuitive / forgivable one in the first case because of the usefulness of approximating partial sums. Writing things as $(1,2,3,...)\sim -\frac{1}{12}$ or $R\left(\right(1,2,3,...\left)\right)=-\frac{1}{12}$  (R() denoting Ramanujan summation, which for convergent series is equivalent to taking the limit of partial sums) would make this all less mysterious.

In other words, (1, 2, 3, ...) is in an equivalence class with -1/12, an equivalence class which also contains any finite series which sum to -1/12, convergent infinite series whose limit of partial sums is -1/12, and divergent series whose Ramanujan sum is -1/12.

It is not true that "no pattern that suggests a value suggests any other", at least not unless you say more precisely what you are willing to count as a pattern.

Here's a template describing the pattern you've used to argue that 1+2+...=-1/12:

We define numbers $a_{ij}$ with the following two properties. First, ${lim}_{j\to \infty }{a}_{ij}=i$, so that for each $j$ we can think of $\left(a_{ij}\right)$ as a sequence that's looking more and more like (1,2,3,...) as $j$ increases. Second, ${\sum }_{i=1}^{\infty }{a}_{ij}=s_j$ where $s_j\to -\frac{1}{12}$, so the sums of these sequences that look more and more like (1,2,3,...) approach -1/12.

(Maybe you mean something more specific by "pattern". You haven't actually said what you mean.)

Well, here are some $a_{ij}$ to consider. When $i>j+1$ we'll let $a_{ij}=0$. When $i\le j$ we'll let $a_{ij}=i$. And when $i=j+1$ we'll let $a_{ij}=A-(1+\cdots +i)$. Here, $A$ is some fixed number; we can choose it to be anything we like.

This array of numbers satisfies our first property: ${lim}_{j\to \infty }{a}_{ij}=i$. Indeed, once $j\ge i$ we have $a_{ij}=i$, and the limit of an eventually-constant sequence is the thing it's eventually constant at.

What about the second property? Well, as you'll readily see I've arranged that for each $j$ we have ${\sum }_{i=1}^{\infty }{a}_{ij}=A$. So the sequence of sums converges to $A$.

In other words, this is a "pattern" that makes the sum equal to $A$. For any value of $A$ we choose.

I believe there are more stringent notions of "pattern" -- stronger requirements on how the $a_{ij}$ approach $i$ for large $j$ -- for which it is true that every "pattern" that yields a finite sum yields $-\frac{1}{12}$. But does this actually end up lower-tech than analytic continuation and the like? I'm not sure it does.

(One version of the relevant theory is described at https://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation.)

I'm one of these professional mathematicians, and I'll say that this article completely fails to demonstrate it's central thesis that there is a valid intuitive argument for concluding that 1 + 2 + 3 + ... = -1/12 makes sense. What's worse, it only pretends to do so by what's essentially a swindle. In my understanding, it's relatively easy to reason that a given divergent series "should" take an arbitrary finite value by the kind of arguments employed here, so what is being done is taking a foregone conclusion and providing some false intuition for why it should be true.

On a less serious note, speaking to the real reason why 1 + 2 + 3 + ... = -1/12, that's actually what physicists will tell you, and we all know one should be careful around those.

The finish was quite a jump for me. I guess I could go and try to stare at your parenthesis and figure it out myself, but mostly I feel somewhat abandoned at that step. I was excited when I found 1, 2, 4, 8... = -1 to be making sense, but that excitement doesn't quite feel sufficient for me to want to decode the relationships between the terms in those two(?) patterns and all the relevant values

Here is another explanation, kind of:

Taylor expansion of 1/(1+x)^2 is 1 - 2x + 3x^2 - 4x^3 + 5x^4...

When x = 1, it means that 1 - 2 + 3 - 4 + 5... = 1/4

But 1 - 2 + 3 - 4 + 5... can be written as 1 + 2 + 3 + 4 + 5... - 2×2 - 2×4 - 2×6...

= 1 + 2 + 3 + 4 + 5... - 2×(2 + 4 + 6...)

= 1 + 2 + 3 + 4 + 5... - 2×2×(1 + 2 + 3...)

= (1 - 2×2) × (1 + 2 + 3...)

= -3 × (1 + 2 + 3...)

So if 1 - 2 + 3 - 4 + 5... = 1/4, we get:

1/4 = -3 × (1 + 2 + 3...)

-1/12 = 1 + 2 + 3...

(Found here.)

Shay you want here is Matthew D.  Schwarz, Quantum Theory and the Standard Model, chapter 15.

(or the original paper by CasiMir cited in the above book chapter.

Zeta regularisation is a much saner way to explain it.

====

in a physics context .. suppose that the laws of physics you have are only valid up to some energy scale E, where E is presumed large.

The physical quantity you 're interested in is f(E) - g(E). 
 

lim E -> infinity of f(E) and g(E) is infinite, so can't safely exchange the order of the limits and the subtraction. But lim E -> infinity (f(E) - g(E)) exists and is finite, so you're good to go, and the result is insensitive to what the energy scale E actually is,

This was a fun read and felt (for me) more simple and follow-able than most things I've read explaining math! Thank you.

I got up to the sums of powers of $2$ being $-1$. That bit took a few close reads but I followed that there was a pattern where infinite sums of $(1,r,r^2,r^3,\dots )$ equal $\frac{1}{1-r}$, and there's reason to believe this holds for $r$ between $-1$ and $1$. Then you write that if we apply it to $r=2$ then it's equal to $-1$, which is a daring question to even ask and also a curious answer! But what justification do you have for thinking that equation holds for $r$ that aren’t between $-1$ and $1$? I think that this was skipped over (though I may be missing something simple).

(Also, if it does hold for numbers that aren't between $-1$ and $1$ that I believe this also implies that all infinite sums of $r^n$ equal to negative numbers, and suggests maybe all infinite sums of positive integers will too.)

If I'm understanding correctly, the argument here is:

A) ${lim}_{x\to \infty }\left({\sum }_{n=1}^{\infty }\right(n{e}^{-\frac{n}{x}}\cos (-\frac{n}{x})\left)\right)=-\frac{1}{12}$

B) ${lim}_{x\to \infty }\left(n{e}^{-\frac{n}{x}}\right)=n$

C) ${lim}_{x\to \infty }\left(\cos \right(-\frac{n}{x}\left)\right)=1$

Therefore, ${\sum }_{n=1}^{\infty }(n\ast 1)=-\frac{1}{12}$.

First off, this seems to have an implicit assumption that ${lim}_{x\to \infty }\left({\sum }_{n=1}^{\infty }\right(f(x,n)\ast g(x,n)\left)\right)={\sum }_{n=1}^{\infty }\left(\right({lim}_{x\to \infty }f(x,n))\ast ({lim}_{x\to \infty }g(x,n)\left)\right)$.

I think this assumption is true for any functions f and g, but I've learned not to always trust my intuitions when it comes to limits and infinity; can anyone else confirm this is true?

Second, A seems to depend on the relative sizes of the infinities, so to speak. If j and k are large but finite numbers, then ${\sum }_{n=1}^j\left(n{e}^{-\frac{n}{k}}\cos \right(-\frac{n}{k}\left)\right)\approx -\frac{1}{12}$ if and only if j is substantially greater than k; if k is close to or larger than j, it becomes much less than or greater than -1/12.

I'm not sure exactly how this works when it comes to infinities - does the infinity on the sum have to be larger than the infinity on the limit for this to hold? I'm pretty sure what I just said was nonsense; is there a non-nonsensical version?

In conclusion, I don't know how infinities work and hope someone else does.