Zeta regularization vs Dirichlet series

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.

Answer

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.