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Suppose you have a sequence of real numbers, denoted $a_n$. Then the sum of the sequence is

$\sum_n a_n$

If this is divergent, we can use zeta regularization to get a sum. We can do this by defining the function

$\zeta_A(s) = \sum_n a_n^{-s}$

and then analytically continue to the case where $s=-1$.

A different approach is to define the Dirichlet series

$A(s) = \sum_n \frac{a_n}{n^s}$

and then analytically continue to the case where $s=0$.

$Questions:$

  1. When these two approaches are both defined, are they guaranteed to agree on the result? If not, for which sequences do they agree?

  2. If they are compatible, is the second summation method strictly stronger than the first?

For instance, it is clear that the first method can't do anything for the series $1+1+1+1+1+...$, whereas the second method yields -1/2, so it is at least as strong as the first.

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  • $\begingroup$ If there is a typo, you can edit your question and fix it. $\endgroup$ – Mariano Suárez-Álvarez Jan 24 '18 at 21:05
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For $n \geqslant 1$, let $a_n = n + (-1)^{n-1}$. Then $a_{2n} = 2n-1$ and $a_{2n-1} = 2n$, so

$$\sum_{n = 1}^{\infty} \frac{1}{a_n^s} = \zeta(s)$$

for $\operatorname{Re} s > 1$. Thus $\zeta$-regularisation leads to $\zeta(-1)$. And for $\operatorname{Re} s > 2$ we have

$$\sum_{n = 1}^{\infty} \frac{a_n}{n^s} = \sum_{n = 1}^{\infty} \frac{1}{n^{s-1}} + \sum_{n = 1}^{\infty} \frac{(-1)^{n-1}}{n^s} = \zeta(s-1) + \eta(s)\,$$

so the analytic continuation of Dirichlet series leads to $\zeta(-1) + \eta(0) = \zeta(-1) + \frac{1}{2}$.

These methods are hence not compatible.

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  • $\begingroup$ Thanks, this is helpful. So they disagree about rearrangements of $a_n$. So then the question is, when do they ever agree? And is there any reason to prefer one method vs the other? $\endgroup$ – Mike Battaglia Jan 25 '18 at 3:24
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    $\begingroup$ They would agree if the rearrangement is such that the resulting Dirichlet series (or its analytic continuation) has a zero at $0$, for example if we rearrange only a finite part and have $a_n = n$ for all large enough $n$. (Yes, that's not very interesting. But I don't know if there are interesting conditions that make them agree. I expect they agree in more interesting cases too, but I don't know any non-obvious examples.) $\zeta$-regularisation dies in the presence of zeros or bounded subsequences, but when it works it is stable under rearrangements, which is a nice property. $\endgroup$ – Daniel Fischer Jan 25 '18 at 15:01
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    $\begingroup$ Analytic continuation of Dirichlet series can work with zeros and bounded subsequences, but is affected by rearrangements (not by finite rearrangements, however). In cases where both methods work, I don't know of a systematic reason to prefer one over the other. $\endgroup$ – Daniel Fischer Jan 25 '18 at 15:01

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