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$\displaystyle\sum_{n=1}^\infty \frac{1}{n^s}$ only converges to $\zeta(s)$ if $\text{Re}(s) > 1$.

Why should analytically continuing to $\zeta(-1)$ give the right answer?

  • 28
    It's not clear (to me) what you mean by "the right answer". Since this was migrated from physics, I suspect that you have in mind the use of this summation in string theory. However, mathematically speaking, there is no "right answer" for summation techniques for divergent series; $-\frac{1}{12}$ is simply the answer that analytical continuation yields. See also http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_·_·_· .2011-05-18
  • 61
    What does the phrase "right answer" mean? If you *define* zeta(s) in the usual way for s with real part greater than 1, it has an analytic continuation to s = -1 and the value there is -1/12. That does not *mean* the series defining zeta(s) for Re(s) > 1 converges at s = -1 with value -1/12 except in a hand-waving Euler kind of way. This question doesn't really have a mathematical meaning as far as I can tell.2011-05-18
  • 19
    See: http://math.stackexchange.com/questions/37327/infinity-1-paradox, http://math.stackexchange.com/questions/27526/s-11010010010000-1-9-how-is-that, and http://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/ for more information too.2011-05-18
  • 7
    Someone should edit the Wikipedia section that says $1+2+3+... = \frac{-1}{12}$ https://en.wikipedia.org/wiki/Riemann_zeta_function#Specific_values2013-12-05
  • 8
    (For newcomers) there are a lot of linked questions now, but most relevant, IMO: i) [on (problems with) finding infinite sums via 'shift & add'](http://math.stackexchange.com/q/37327/), ii) [on analytic continuation](http://math.stackexchange.com/questions/41938/), iii) [on assigning values to divergent series](http://math.stackexchange.com/q/632985/)2014-01-12
  • 2
    See also [Infinite series are weird](http://skullsinthestars.com/2010/05/25/infinite-series-are-weird-redux/) blog post2014-01-19
  • 1
    Watch this video: http://youtu.be/wt6ngy6pDws2014-04-08
  • 0
    https://www.youtube.com/watch?v=w-I6XTVZXww2015-01-20
  • 1
    I would classify the result as an eigenvalue. The summation behaves as if it were this value in specific cases, however the properties, operations and definitions of the summation are not limited to those behaviors. It, in fact, breaks axiom of closure, if we consider each term of the sum as an element in the natural numbers -- since we fail to obtain a result that is an element in the naturals, just as with matrices. There are probably other similarities that could be shown as well, but I'm being fairly lazy today.2015-05-06
  • 2
    You can check out :https://www.youtube.com/watch?v=0Oazb7IWzbA2015-05-26
  • 0
    For those people who agree with $1-1+1-1+\cdots=\frac12$: If this is true then according to you'll $\lim\limits_{x\to\infty} \sin(x)=$ [Average of all possible values of $\sin(x)$]=$\frac1{\sqrt{2}}$? (Because you say that the sum converges to half because it is the average of possible values of it)2015-08-20
  • 0
    By average of all possible values I mean all positive, if we take negative also, then $f_{avg}=0$ for $\sin(x)$.2015-08-20
  • 1
    Re the three videos linked to in the comments above, [the one by MrYouMath](https://www.youtube.com/watch?v=wt6ngy6pDws) is to be commended while the first Numberphile video is, for its main part, a lazy exercise of self complacency, which caused enough reactions to motivate its authors to post the other video linked above, as a kind of (not very convincing, if you ask me) damage control operation.2016-05-02
  • 0
    @JustKevin The moment we tried to evaluate $S=1+2+3+\dots$, it broke the axiom of logic (and I guess it broke perturbation theory). :'(2016-09-11
  • 0
    @AdityaAgarwal Nah, by the MVT and the fact that $\int_0^N\sin(x)dx$ is bounded, the average value of $\sin$ for positive values is $0$, in the limit sense.2016-09-11
  • 1
    @Did Agreed. I feel Mathologer's video is also enlightening to the layman in at least the dangers of such layman methods while noting at more acceptable ways. https://www.youtube.com/watch?v=jcKRGpMiVTw2016-09-11
  • 1
    @SimpleArt Indeed, there is a stark contrast between Mathologer's careful, useful, informative approach and Numberphile's nonsense. And one simply has to love how Mathologer very briefly mentions Numberphile's piece at 1:46-1:55... so delicately put! (See also the first comment below the video.) Thanks for mentioning Mathologer's work.2016-09-11
  • 0
    Take a look at [Ramanujan summation article at wikipedia](http://en.wikipedia.org/wiki/Ramanujan_summation).2011-07-12

15 Answers 15