

1 + 2 + 3 + 4 + ⋯ = -1/12 - baby
http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF

======
ColinWright
The better to understand the ideas behind this, first it's useful to
understand why, using the same reasoning, we have:

    
    
      1 + 2 + 4 + 8 + 16 + 32 + ... = -1
    

You can read about that here[0], and I've submitted that as a separate link
here[1].

It's interesting that under twos-complement arithmetic it is, in fact, also
true.

[0]
[http://www.solipsys.co.uk/new/BeyondTheBoundary.html?HN20150...](http://www.solipsys.co.uk/new/BeyondTheBoundary.html?HN20150106)

[1]
[https://news.ycombinator.com/item?id=8843372](https://news.ycombinator.com/item?id=8843372)

------
eximius
No! Just NO!

For the love of God and everything holy, it does not _equal_ -1/12, it is just
a value we use in place of its divergent value.

~~~
ColinWright
It's more than "... value we use in place of its divergent value." It's the
value at _s=-1_ of the analytic continuation of the series:

    
    
        \zeta(s) = sum_{n=0}_{oo} n^-s
    

In other words, we take a series that converges for some values, and then see
if it has an analytic continuation. In this case it does, so we look at the
value of the analytic continuation at places where the original series is not
defined.

As a simpler example, consider 1+x+x^2+x^3+x^4+... This is not defined outside
of the disk |z|<1\. However, when |z|<1 we have:

    
    
       1/(1-x) = 1+x+x^2+x^3+x^4+...
    

So for x=2 we can evaluate 1/(1-x), but we can't evaluate the infinite series.
When x=2 the value of 1/(1-x) is -1, does that mean we can say that
1+2+2^2+2^3+2^4+2^8+... is -1? Well, no, but sort of.

Playing fast and loose with these things is almost guaranteed to go wrong, but
when you _really_ understand what's happening, saying things like:

    
    
        1+2+3+4+5+6+... = -1/12
    

... is a useful shorthand, and justifiable.

