Can different choices of regulator assign different values to the same divergent series?

Question

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?

Answer 1

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 $g(0)=\beta$.