Suppose $\sum_{n=1}^{\infty}a_n$ is a divergent sum. Define its power series regularized value ($\sf P$) to be the analytic continuation of $\sum_{n=1}^{\infty} a_n z^n$ evaluated at $z=1$, and its Dirichlet series regularized value ($\sf D$) to be the analytic continuation of $\sum_{n=1}^{\infty} a_nn^{-s}$ evaluated at $s=0$.
Conjecture. If the power series regularized value of a divergent sum exists, then so does the Dirichlet series regularized value, and they are equal.
Is this true? Or known to be true for a certain class of divergent sums?
My only evidence for this comes from the Riemann zeta function. We have
$$ \begin{array}{lllll} 1+1+1+\cdots & = & \zeta(0) & = & -\frac{1}{2} \\ 1+2+3+\cdots & = & \zeta(-1) & = & -\frac{1}{12} \end{array} \tag{$\sf D$} $$
however neither of these have a power series regularized values, since
$$ \begin{array}{lllll} 1+1+1+\cdots & = & \frac{1}{1-1} & = & \infty \\ 1+2+3+\cdots & = & \frac{1}{(1-1)^2} & = & \infty. \end{array} \tag{$\sf P$}$$
This can be remedied by looking at alternating series:
$$ \begin{array}{llllcll} 1-1+1-\cdots & = & \eta(0) & = & (1-2^{1-0})\zeta(0) & = & \frac{1}{2} \\ 1-2+3-\cdots & = & \eta(-1) & = & (1-2^{1-(-1)})\zeta(-1) & = & \frac{1}{4} \end{array} \tag{$\sf D$}$$
where $\eta(s)=\sum_{n=1}^{\infty}(-1)^{n-1}n^{-s}=(1-2^{1-s})\zeta(s)$ is the Dirichlet eta function, and
$$ \begin{array}{lllll} 1-1+1-\cdots & = & \frac{1}{1-(-1)} & = & \frac{1}{2} \\ 1-2+3-\cdots & = & \frac{1}{(1-(-1))^2} & = & \frac{1}{4} \end{array} \tag{$\sf P$}$$
are the power series regularized values. More generally,
$$ \frac{k!}{(1-w)^{k+1}} = \sum_{n=1}^{\infty} n(n-1)\cdots(n-(k-1))w^{n-k} $$
(after differentiating $(1-z)^{-1}$ a total of $k$-times), which gives
$$ \frac{k!}{(1-z)^{k+1}}= \sum_{n=1}^{\infty} \left(\sum_{r=0}^k s(k,r)n^r \right)w^{n-k} $$
$$ \frac{k!w^k}{(1-w)^{k+1}} = \sum_{r=0}^k s(k,r) \mathrm{Li}_{-r}(w). $$
(Note $s(k,r)$ are the Stirling numbers.)
This can be used as a valid analytic continuation of $\mathrm{Li}_{-r}(w)$ to specialize e.g. $w=-1$ which ought to generate the previous observations with $\mathrm{Li}_{-r}(-1)=-\eta(-r)$.
Best Answer
The power series summation of $(-2)^n$ is well-defined but it doesn't have a Dirichlet series summation.
If the Dirichlet series $F(s)=\sum_{n=1}^\infty a_nn^{-s}$ converges for some $s_0$ then $a_n = O(n^{s_0})$,
$F(s)$ converges absolutely and it is analytic for $\Re(s)> \Re(s_0)+1$,
$f(z) = \sum_{n=1}^\infty a_nz^n$ is analytic for $|z| < 1$ and $$\Gamma(s)F(s) = \int_0^\infty t^{s-1} f(e^{-t})dt, \qquad \Re(s) > \Re(s_0)+1$$ If also $f(1)=\lim_{z \to 1^-}f(z) $ exists then the latter integral converges and is anaytic for $\Re(s) > 0$. Moreover we have $$\Gamma(s)F(s) =f(1) \Gamma(s)+ \int_0^\infty t^{s-1} (f(e^{-t})-f(1)e^{-t})dt, \qquad \Re(s) > 0$$ where $$f(e^{-t})\!-\!f(1)e^{-t} = o(1) \implies \int_0^\infty t^{s-1} (f(e^{-t})-f(1)e^{-t})dt = o(\frac1{\Re(s)}) \implies f(1) = \lim_{s \to 0^+} F(s)$$ If also $\lim_{z \to 1^-}f'(z) =f'(1)$ exists then $f(e^{-t})-f(1)e^{-t} = O(t)$ so that $\int_0^\infty t^{s-1} (f(e^{-t})-f(1)e^{-t})dt$ converges and it is analytic for $\Re(s) > -1$ which means $F(s)$ can be analytically continued to $\Re(s) > -1$ and hence