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How does Dirichlet regularization assign value $-\frac{1}{12}$ to $\sum_{k=1}^{\infty} k$?

Yes, I know that $\zeta(-1) = - \frac{1}{12}$, a result that follows from the Riemann functional equation $\zeta(s) = 2^s \pi^{s-1} \sin(\frac{\pi s}{2})\, \Gamma(1 - s)\, \zeta(1-s)$.

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murray
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    Well, the fact of the matter is that it simply isn't. You can't regularise $\sum_{k=1}^{\infty }k$ with such a construction: the sum isn't even defined for $\Re{s}<2$, let alone a neighbourhood of zero. – Chappers Oct 03 '15 at 00:34
  • OK, then I rephrase: just what series is the Dirichlet regularization of $\sum_{k=1}^{\infty} 1/k^s$ and why is the value of that regularization (not the original series itself, of course) equal to $-1/12$ when $s =-1$? – murray Oct 03 '15 at 00:43
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    https://en.wikipedia.org/wiki/Analytic_continuation To get the full picture, look at formulae for the zeta function: https://en.wikipedia.org/wiki/Riemann_zeta_function Then show that they agree on $ \Re z > 1$, analytic continuation, baby! You can get a series if you want by knowing that any holomorphic function on an open set can be expanded as a power series. – Almentoe Oct 03 '15 at 01:01
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    I think this question would be better phrased as "How do you show that $\zeta(-1)=-1/12$?" – Mario Carneiro Oct 03 '15 at 07:19
  • @Almentos: Which particular formula do you refer to that I should see agrees with $\sum_{k=1}^{\infty} k^{-s}$ when $\Re(s) > 1$? – murray Oct 08 '15 at 15:40
  • You know that $\zeta(-1)=-1/12$, so... why are you asking a question you know the answer to? – anon Oct 08 '15 at 16:13
  • I'm trying to understand exactly what one does to the series $\sum_{k=1}^{\infty} k$ so as to produce a series having a parameter $s$ form which some kind of limit produces value $-1/12$. – murray Oct 08 '15 at 16:16
  • The zeta regularization of the series $\sum_{n=1}^\infty a_n$ is attained by analytically continuing the function $\sum_{n=1}^\infty a_n^{-s}$ and then evaluating the resulting function at $s=-1$. – anon Oct 08 '15 at 16:37
  • But just how does one analytically continue $\sum_{n=1}^{\infty} a_n^{-s}$, that is, how does one obtain a formula that one can then evaluate at $s=-1$? – murray Oct 08 '15 at 18:31
  • In the case of $a_n:=n$... by using the functional equation, which you gave in your question. – anon Oct 08 '15 at 23:24
  • That does not answer my question: what alteration does one make to the terms of the series, and what limit is then taken (and how) -- by the method of DIRICHLET REGULARIZATION -- so as to assign a value to the "sum" of the given divergent series? – murray Oct 10 '15 at 19:06
  • @whacka: OK, but why does the analytic continuation of $\sum_{n=1}^{\infty} n^{-s}$ have value $-1/12$ at $s = -1$? Just saying it does have this value there does not explain why it is so! – murray Jul 02 '21 at 18:52

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$$\sum_{n=1}^{\infty}c^n *n$$ look around c=-1 and you got your answer of the dirichlet sum. (-1/4)

$$\sum_{n=1}^{dp} f(n)\sum_{k=1}^{d-1} (e^{\frac{2i\pi k}{d}})^{n}=\sum^p_{n=1} (d f(nd)-f(n))$$

Now we look around the limits here, given $f(n)=n$. We see at d=2 (aka c=-1 above), that it will be equal 3 times $\zeta(-1)$ so $\zeta(-1)=-1/12$

Gerben
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  • How precisely do you get the value of $\sum_{n=1}^{\infty} n,c^n$ for values of $c$ that are around 1? – murray Jan 19 '16 at 20:35
  • What's the relevance of the 2nd formula that begins with the double sum $\sum{n=1}^{d,p} f(n),\sum_{k=1}^{d-1}(e^{2i\pi k/d})^n$? And what is the purpose of that double sum — is taking a limit of that the very definition of Dirichlet regularization of the series $\sum_{n=1}^{\infty} f(n)$? Please help me by being precise and even prolix in your answer! – murray Jan 21 '16 at 17:07
  • I tried to write it down in english. If your willing to correct the story, i'll post it. It's informal but might explain one and other.But it turns out it's about 2-3 pages long. I could just mail it to you if you wish. You find $\sum_{n=1}^{infty}n$ by looking at the geometric series, which got a pole at $c=1$. That's why you need to esteblish a relation ship for another value and $c=1$. This is what the second formula does and why it's relevant. – Gerben Jan 25 '16 at 17:48
  • Thank you, I'd like to see that explanation. My email address appears on: http://www.math.umass.edu/directory/emeritus-faculty/murray-eisenberg – murray Jan 25 '16 at 21:35
  • I have send you the document. As stated i'm far from skilled engough to make any claim on it being the correct way. I was just as intrigued as you and tried to create an explaination. But it's better then nothing,I took a shot. Turned out to be 5 pages long with way too much talk on my side, which is why i won't post it here. – Gerben Jan 27 '16 at 17:29