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

anon
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  • where is $\sum_n a_nz^n$ analytic? you have to give more information.... – mathworker21 Aug 20 '19 at 04:28
  • @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 $z=0$. – anon Aug 20 '19 at 04:32
  • 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 $(a_n)_{n\in\mathbb{N}}$. – Jfischer Aug 20 '19 at 06:55
  • @Jfischer Divergent means not convergent. – anon Aug 20 '19 at 06:59
  • @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 $-\infty$. – Jfischer Aug 20 '19 at 07:11
  • @Jfischer Why does it matter? The series is supposed to be arbitrary. If $\sum_{n=1}^{\infty} a_nz^n$ defines an analytic function which can be analytically continued to an open domain containing $1$ (which requires the original radius of convergence to be $>0$) , then the value of the function at $z=1$ will be the power series regularized value. If the radius of convergence of $\sum_{n=1}^{\infty} a_nz^n$ is $0$, as in your example, then $\sum_{n=1}^{\infty} a_n$ simply does not have a power series regularized value. – anon Aug 20 '19 at 07:16
  • are the $a_n$ real numbers? – Jfischer Aug 20 '19 at 09:39
  • @runway44 really? do you know the difference between a statement and its converse? saying "if it is analytic, it would be analytic around 0" is 100% irrelevant. you need to know it is analytic around 0, which you don't necessarily know. – mathworker21 Aug 20 '19 at 13:45
  • @mathworker21 I agree it is irrelevant. That's why I asked "Why does it matter?" When you say "you need to know [the power series] is analytic around $0$," if you mean it needs to be analytic around $0$ as a necessary (but not sufficient) condition for the power series regularized value to exist, then I agree - and again, I already stated that in my previous comment, and I thought it was obvious from my original question anyway. But if you mean something else by your last sentence, then I don't know what you mean. (As for Jfischer's recent comment: I have not said we're assuming they're real.) – anon Aug 21 '19 at 02:09

1 Answers1

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The power series summation of $(-2)^n$ is well-defined but it doesn't have a Dirichlet series summation.

The main relation between power series and Dirichlet series summation is thisone

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.

reuns
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  • To be clear, this means power series regularization and Dirichlet series regularization are compatible, but neither encompasses the other? – user76284 Dec 14 '22 at 18:16