Timeline for Does any method of summing divergent series work on the harmonic series?
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 3, 2019 at 8:35 | comment | added | Anixx | @user76284 yes, my typo. | |
| Jul 3, 2019 at 0:43 | comment | added | user76284 | @Anixx Since the zeta function starts from $k=1$, shouldn't that be $\sum_{k=1}^\infty 1 = \zeta(0) = -1/2$ and $\sum_{k=0}^\infty 1 = 1 + \zeta(0) = 1 - 1/2 = 1/2$? | |
| Sep 29, 2017 at 22:42 | comment | added | reuns | @DavidSpeyer $S = 1+S$ is what you get only if the summation is shift invariant, which is not the case for the zeta summation (that's why it obtains a finite value $-1/2$) | |
| Sep 4, 2017 at 20:59 | comment | added | Anixx | $\sum_{k=0}^\infty 1 = -1/2$, $\sum_{k=1}^\infty 1 = 1/2$... | |
| Aug 16, 2011 at 20:31 | comment | added | Andrew | By the way the proposition that $1+1+1+... = -1/2$ was first made by Euler. | |
| Oct 29, 2009 at 12:27 | comment | added | David E Speyer | Hmmm. Perhaps I was too strong there. I tend to assume that a summation method should obey a_1+a_2+...=0+a_1+a_2+..., which zeta regularization does not. But I can't make a strong argument for that assumption. | |
| Oct 29, 2009 at 12:23 | comment | added | Armin Straub | You don't think that 1+1+1+... = -1/2 has some decency? (Of course, that's \zeta(0).) | |
| Oct 29, 2009 at 10:40 | history | answered | David E Speyer | CC BY-SA 2.5 |