Is regularization of infinite sums by analytic continuation unique?

Question

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?

Answer 1

Let $$f_k(s) = k^{-s}+(s+1)k^{-s-2},\qquad f_k(-1)=k$$ then $$F(s)=\sum_k f_k(s) = \zeta(s)+(s+1)\zeta(s+2), \qquad F(-1)=-1/12+1$$

Answer 2

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}

Answer 3

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.