# Divergent sums via analytic continuation: power series vs Dirichlet series

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 $$\mathrm{Li}_{-r}(-1)=-\eta(-r)$$.

• where is $\sum_n a_nz^n$ analytic? you have to give more information.... Aug 20, 2019 at 4:28
• @mathworker21 Since it's a power series around $z=0$, I expect if it defines an analytic function at all it would have to be analytic around $z=0$. Aug 20, 2019 at 4:32
• Do you mean by divergent, divergent to infinity or just not a defined limit? Also, assuming that an analytic continuation exists should enforce properties of $(a_n)_{n\in\mathbb{N}}$. Aug 20, 2019 at 6:55
• @Jfischer Divergent means not convergent. Aug 20, 2019 at 6:59
• @runway44 Obviously, but which kind of the two I described? Do you know that $\sum_{n} a_n = \pm\infty$? Consider for example $a_n = (-1)^n n!$. Then the radius of convergence of the power series is $0$ but also $\sum_{n} a_n$ does not tend to either $+\infty$ or $-\infty$. Aug 20, 2019 at 7:11

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
The power series summation and the Dirichlet series summation of $$a_n$$ are equal.