3 ms·

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.

• zaroth 12y ago
  ...I told him that the sum of an infinite number of terms of the series: 1 + 2 + 3 + 4 + · · · = −1/12 under my theory. If I tell you this you will at once point out to me the lunatic asylum as my goal. I dilate on this simply to convince you that you will not be able to follow my methods of proof if I indicate the lines on which I proceed in a single letter.... -- Ramanujan
• ColinWright 12y ago
  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.