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Physicists often assign a finite value to a divergent series $\sum_{n=0}^\infty a_n$ via the following regularization scheme: they find a sequence of analytic functions $f_n(z)$ such that $f_n(0) = a_n$ and $g(z) := \sum_{n=0}^\infty f_n(z)$ converges for $z$ in some open set $U$ (which does not contain 0, or else $\sum_{n=0}^\infty a_n$ would converge), then analytically continue $g(z)$ to $z=0$ and assign $\sum_{n=0}^\infty a_n$ the value $g(0)$. Does this prescription always yield a unique finite answer, or do there exist two different sets of regularization functions $f_n(z)$ and $h_n(z)$ that agree at $z=0$, such that applying the analytic continuation procedure above to $f_n(z)$ and to $h_n(z)$ yields two different, finite values?

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    $\begingroup$ Question reopened. $\endgroup$ – Jack D'Aurizio Jul 10 '16 at 6:16
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    $\begingroup$ I've also heard a different resolution, which is that when regulators disagree, the 'true' answer is zeta regulatization. I have no idea why nature would favor $\zeta$, though. $\endgroup$ – knzhou Jul 10 '16 at 17:44
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    $\begingroup$ @knzhou There are definitely many examples (including yours) where one choice of regulator gives a finite answer and another gives an infinite answer. This case doesn't bother me as much because it's natural to favor the regulator that gives a finite answer. I'm more concerned with two regulators that yield different finite values, because then it really does seem unclear which one would be physically correct. $\endgroup$ – tparker Jul 10 '16 at 19:57
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    $\begingroup$ @tparker Right, my question is really about comparing analytic continuation to the other procedure physicists use, which is to parametrize the divergence and subtract it out. But my intuition goes the exact opposite way from yours; I trust the latter since I can see what it's doing to the UV. I don't see what analytic continuation is doing. $\endgroup$ – knzhou Jul 10 '16 at 20:08
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    $\begingroup$ @tparker If you feel that your question hasn't received enough attention in the next couple days, let me know. I feel it's a fine and interesting one, and wouldn't mind putting a bounty on it if it can help. $\endgroup$ – Clement C. Jul 11 '16 at 15:55
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The way you have the question written, the procedure can absolutely lead to different, finite, results depending on one's choice of the $f_n(z)$. Take the simple example of $1-1+1-1+\ldots$. The most obvious possibility is to take $f_n(z)=\frac{(-1)^n}{(z+1)^n}$ (i.e., a geometric series), in which case $$ g(z)=\sum_{n=0}^{\infty}f_n(z)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(z+1)^{n}}=\frac{1}{1+\frac{1}{z+1}}=\frac{z+1}{z+2}, $$ where the sum converges for $|z+1|>1$, and $g(0)=1/2$. But if you don't insist on the terms forming a power series, then there are other possibilities. For instance, let $f_{2m}(z)=(m+1)^z$ and $f_{2m+1}(z)=-(m+1)^z$ (i.e., zeta-regularize the positive and negative terms separately); then $g(z)=0$ everywhere, where the sum converges for $\Re(z) < -1$ and is analytically continued to $z=0$.

By taking an appropriate linear combination of the first and second possibilities, you can get $1-1+1-1+\ldots$ to equal any value at all. Specifically, taking $$ f_n(z)=(-1)^n \left(\frac{2\beta}{(z+1)^n}+(1-2\beta)\left\lceil\frac{n+1}{2}\right\rceil^z\right), $$ you find $g(z)=2\beta(z+1)/(z+2)$, convergent in an open region of the left half-plane, and $g(0)=\beta$.

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    $\begingroup$ see also this question 'non-canonicity-of-using-zeta-function-to-assign-values-to-divergent-series' $\endgroup$ – reuns Jul 18 '16 at 14:19
  • $\begingroup$ Great answer. Is there an example where $f_n(z)$ and $h_n(z)$ are both power series? $\endgroup$ – tparker Jul 18 '16 at 16:58
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    $\begingroup$ @tparker: If you're looking at a Taylor series around some point $z_0$, with a nonzero radius of convergence, then the analytic continuation of $g(z)$ from $z_0$ to $0$ may depend on the choice of continuation path (e.g., if there are branch cuts). But other than the choice of branch, I believe the result is uniquely defined with the constraint that $f_n(z)=a_n (z-z_0)^n$. $\endgroup$ – mjqxxxx Jul 18 '16 at 17:49
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    $\begingroup$ In other words, if you choose $z_0$, and then find the Taylor series around $z_0$ that matches the given values at $0$ ($f_n(z)=a_n (-z_0)^{-n}(z-z_0)^n=a_n(1-z/z_0)^n$), then you get the same function up to dilation around $z=0$, and the value of $g(0)$ is invariant under the choice of $z_0$. $\endgroup$ – mjqxxxx Jul 18 '16 at 20:11
  • $\begingroup$ There's something very special about $\beta = 1/2$ though: It's the only one for which $\sum_{n=0}^\infty f_n(z)$ converges near $z=0$ (in fact, for all $z>0$)! imgur.com/a/Ln0a9L0 $\endgroup$ – user76284 Jul 10 at 3:27

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