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There are ill-posed summations that we can assign values to, take for concreteness, $$ S = \sum_{k=0}^\infty k $$ to which we can assign $-1/12$ by several methods. Is there a fundamental and rigorous reason why these methods have to agree on the same value?

Perhaps the most powerful means we have available is the zeta function regularization, where we attach $\zeta(-1)$ (Riemann zeta function), which formally represents $S$, to the summation. If we have another analytic function that formally represents $S$ and analytically continues to assign a value to $S$, must the value always be the same, $-1/12$?

More concretely, suppose I have a function $\xi(s) = \sum_{k=0}^\infty f_s(k)$ such that $f_1(k) = k$ and $f_s(k)$ is a "nice" function (for the case of zeta function, $f_s(k)=k^{-s}$ is the exponential function in $s$). Suppose that $\xi(s)$ can be analytically continued to a meromorphic function with a value at $s=-1$. Must it be the case that $\xi(-1)=-1/12$?

More generally, what can be said about other divergent sums? If there is an analytic continuation that defines the value of such sums, must it be unique? Where can I look up known uniqueness results and techniques?

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    $\begingroup$ For your example $f_s(k)=k^{-s}$ we have $f_1(k)=1/k$, not $k$. $\endgroup$ – Richard Stanley May 11 at 22:11
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    $\begingroup$ Let $\xi(s)=\sum_{k\geq 1}ke^{-(s+1)k}$. This extends to a meromorphic function on $\mathbb{C}$ for which $s=-1$ is pole, rather than $\xi(-1)=-1/12$. $\endgroup$ – Richard Stanley May 12 at 3:08
  • $\begingroup$ Well, we can add the requirement that the function is defined to have a finite value (no poles) at s=-1. $\endgroup$ – MCH May 13 at 3:45
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Let $$f_k(s) = k^{-s}+(s+1)k^{-s-2},\qquad f_k(-1)=k$$ then $$F(s)=\sum_k f_k(s) = \zeta(s)+(s+1)\zeta(s+2), \qquad F(-1)=-1/12+1$$

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  • $\begingroup$ It seems to me that $F(-1)=-1/12$. $\endgroup$ – Richard Stanley May 12 at 14:15
  • $\begingroup$ $\lim_{s\to 1}(s-1)\zeta(s)=1$ $\endgroup$ – reuns May 12 at 16:44
  • $\begingroup$ Oops, you are right. $\endgroup$ – Richard Stanley May 12 at 17:47
  • $\begingroup$ Mathematica tells me that Richard Stanley is right... $\endgroup$ – MCH May 12 at 21:13
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    $\begingroup$ @MCH Interesting point. Reminds me of physics.stackexchange.com/a/492793/35699. Perhaps a stronger condition is implicitly lurking here. $\endgroup$ – user76284 May 13 at 19:18
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As the other answer has pointed out, $-1/12$ is not the only value that can obtained with analytic continuation. However, it is the unique constant term of the asymptotic expansion of the smoothed partial sums, which perhaps explains why it is the most "natural" value.

Let $\eta$ be any Schwartz function such that $\eta(0) = 1$. Then

\begin{align} \sum_{n=1}^\infty n^s \eta(n \varepsilon) &= \zeta(-s) + O(\varepsilon) + \frac{1}{\varepsilon^{s+1}} \int_0^\infty x^s \eta(x) dx \end{align}

Therefore, by choosing for any given $s$ an $\eta$ that makes the last integral zero, we get

\begin{align} \sum_{n=1}^\infty n^s &= \sum_{n=1}^\infty n^s \lim_{\varepsilon \rightarrow 0^+} \eta(n \varepsilon) \\ &\overset{!}{=} \lim_{\varepsilon \rightarrow 0^+} \sum_{n=1}^\infty n^s \eta(n \varepsilon) \\ &= \lim_{\varepsilon \rightarrow 0^+} \left( \zeta(-s) + O(\varepsilon) \right) \\ &= \zeta(-s) \end{align}

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  • $\begingroup$ Is it obvious an $\eta(x)$ that makes that integral vanish exists for all $s$ independently of $s$? $\endgroup$ – AlexArvanitakis May 13 at 3:21
  • $\begingroup$ @AlexArvanitakis I found a class of functions here that seem to cover all $s \neq -1$, namely $\eta_s(x) = \exp(-x)\left(1 - \frac{x}{s+1}\right)$. $\endgroup$ – user76284 May 13 at 3:28
  • $\begingroup$ That won't work for you; the $\eta$ on the LHS is $s$-independent. $\endgroup$ – AlexArvanitakis May 13 at 3:29
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    $\begingroup$ @AlexArvanitakis The series is solved independently for each $s$. I've edited the post to clarify. $\endgroup$ – user76284 May 13 at 3:30
  • $\begingroup$ Thanks, that helped me $\endgroup$ – AlexArvanitakis May 13 at 14:22
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As the other answer showed, the regularization by analytic continuation is not unique.

The problem is mostly in that you choose this method. If you took other methods, like Dirichlet or Borel regularization, in most cases you would obtain the same result.

Analytic continuation in general does not belong here. Maybe there can be some constraints on this method that would restrict its results to those which are obtained by other methods, but so far it seems nobody proposed any.

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    $\begingroup$ What is 'here' in "Analytic continuation in general does not belong here"? $\endgroup$ – LSpice May 12 at 18:39
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    $\begingroup$ @LSpice analytic continuation does not belong to the set of the mutually-compatible methods that produce the same result when applicable (Abel, Cesaro, Borel, Dirichlet, etc). Analytic continuation without additional restrictions can yeld, in general, any result. $\endgroup$ – Anixx May 12 at 18:43
  • $\begingroup$ Thanks. I understand that different transformations that assign values to infinite sums (Borel sums etc) may result in different values. But I'm curious to know if, for example for the sum S above, there are possible functions (other than Riemann zeta) that can be analytically continued to a value different from -1/12. $\endgroup$ – MCH May 12 at 21:11
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    $\begingroup$ @The thing is, the most of regularization methods AGREE on the same value. Alanytic continuation is less rigorous in this respect and I am sure you can find such function that it will give any value you want. $\endgroup$ – Anixx May 12 at 21:16
  • $\begingroup$ @Anixx At first the idea is that $\sum^L (-1)^n = \lim_{z\to 1} \sum_{n\ge 0} (-1)^n z^n$ is not the same as $\sum^A (-2)^n = \frac1{1+2z}|_{z=1}$. The former is a limit regularization and has good chances to be related from one method to the other while the latter is analytic continuation and usually is not related (except for zeta/power series with finitely many poles because one is the Mellin transform of the other). $\endgroup$ – reuns May 12 at 23:45

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