I was working with some Dirichlet series and I realized that I have never seen any general conditions under which

\begin{equation} \sum_{n=1}^{\infty}\frac{a_n}{n}=\lim_{s\to1^+}\sum_{n=1}^{\infty}\frac{a_n}{n^s}\label{1}\tag{1} \end{equation}

holds. This is obviously not true in a general case since if so there would be a very simple proof of the PNT from just applying this to $a_n=\mu(n)$. My question is: under what conditions does \eqref{1} hold?

I can show that if $\sum_{n=1}^{\infty}\frac{a_n}{n}$ converges then \eqref{1} must hold using a very simple proof, but I can't find any broader statements. The ideal condition that I would like to show is that if the partial sums $\sum_{n=1}^{N}\frac{a_n}{n}$ are bounded then \eqref{1} must hold. I don't know how I would go about proving this though, and any insights on this general area would be greatly appreciated.

  • 8
    $\begingroup$ If you take $a_n=\frac{\Lambda(n)}{n^{it}}$ for a fixed $t \ne 0$, RHS converges to $-\zeta'(1+it)/\zeta(1+it)$ but LHS oscillates finitely (as $LHS_N-RHS$ ~$-N^{-it}/{it}$) so the partial sums are bounded but the series doesn't converge; for real $a_n$ Hardy-Littlewood proved that $a_n \ge -C/\log n$ is a sufficient condition for 1 above to hold - conditions like this are part of Tauberian theory for which there is a standard reference by J Korevaar - springer.com/gp/book/9783540210580 where you will find lots of stuff about this $\endgroup$
    – Conrad
    Jun 9, 2020 at 2:45

2 Answers 2


Here are analogous Tauberian theorems for power series and Dirichlet series that involve a condition of analytic continuation to a boundary point plus one extra condition that is necessary for the series to converge at some point on that boundary.

Power series: If $c_n \to 0$ then $\sum_{n \geq 0} c_nz^n$ converges for $|z| < 1$. Fatou showed this series converges at each number $z$ with $|z| = 1$ to which the series admits an analytic continuation from inside the disc $\{z : |z| < 1\}$. Note that "$c_n \to 0"$ is a necessary condition for $\sum c_nz^n$ to converge at a number $z$ where $|z| = 1$, so including it as a hypothesis is not unreasonable.

Dirichlet series: If $(a_1 + \cdots + a_n)/n \to 0$ then $\sum a_n/n^s$ converges for ${\rm Re}(s) > 1$. Riesz showed this series converges at each $s$ with ${\rm Re}(s) = 1$ to which the series admits an analytic continuation from the half-plane ${\rm Re}(s) > 1$. The condition $(a_1 + \cdots + a_n)/n \to 0$ is necessary for $\sum a_n/n^s$ to converge at a number $s$ where ${\rm Re}(s) = 1$, so including that as a hypothesis is not unreasonable.

For your motivating example with $a_n = \mu(n)$, of course the hypothesis $(a_1 + \cdots + a_n)/n \to 0$ is known by completely elementary methods to be equivalent to the Prime Number Theorem (in the form "$M(x) = o(x)$" for $M(x) = \sum_{m \leq x} \mu(m)$), so this theorem of Riesz isn't really a good approach to proving the Prime Number Theorem even though the analytic continuation hypothesis for $\sum \mu(n)/n^s = 1/\zeta(s)$ from ${\rm Re}(s) > 1$ to ${\rm Re}(s) = 1$ is valid.

  • $\begingroup$ I'm not sure if the answer to this is known, but would it be sufficient to show that $\liminf_{n\to\infty}(a_1+...+a_n)/n\to0$ for the interchange to be valid? $\endgroup$
    – Milo Moses
    Jun 18, 2020 at 18:18
  • $\begingroup$ I don't know, but I think it is unrealistic to think you're going to find a very simple proof of the Prime Number Theorem this way. Even if a simple-sounding condition is sufficient, surely the proof of that is going to involve serious work or the verification that an example actually fits the condition is going to involve serious work. $\endgroup$
    – KConrad
    Jun 18, 2020 at 19:22
  • $\begingroup$ Actually, the main example I have in mind is not proving the prime number theorem (the methods I am using have already led me to a new all-elementary proof of the PNT). I am actually trying to prove a different conjecture of mine which would follow nicely from a simple condition to let me interchange limit. If you want to discuss this further I would love to, and we can move this conversation to a private chat. $\endgroup$
    – Milo Moses
    Jun 18, 2020 at 19:45
  • $\begingroup$ If you are genuinely sure you have a new all-elementary proof of PNT, try to reach out to one of the number theorists near you (well, could be hard to meet "in person" these days). All conditions you have wished to be true here so far are not true, so you had better be really sure your proof of PNT does not involve a subtle misunderstanding of what is legitimate with limit processes. In particular, simple conditions that justify interchanging of limits could themselves have proofs as hard as that of PNT! See Korevaar's book on Tauberian theorems for a survey of what is known in this area. $\endgroup$
    – KConrad
    Jun 18, 2020 at 20:00
  • $\begingroup$ I made sure that all interchanges of limits were very justified, and I used no high-level theorems. It is not a full proof of the prime number theorem since the approach taken was showing that the average order of the Mobius function is zero but I only showed that if it must exist it must be zero without showing that it exists (I actually prove something stronger but that's the gist of it). Do you wish to discuss this further? I do not know who to reach out to about it. $\endgroup$
    – Milo Moses
    Jun 18, 2020 at 20:38

In D. J. Newman's paper

A simple analytic proof of the prime number theorem

published in The American Mathematical Monthly 87 (1980) 693-696, Newman proved a result of this type which may also be useful to you (this is a minor variant of the criterion given by KConrad in his answer).

It says the following:

Let $|a_n| \leq 1$ and suppose that the Dirichlet series

$$\sum_{n = 1}^\infty \frac{a_n}{n^s}$$

admits a holomorphic continuation to the line $\mathrm{re}\, s = 1$. Then

$$\sum_{n = 1}^\infty \frac{a_n}{n^s}$$ converges for all $\mathrm{re}\, s \geq 1$.

  • 1
    $\begingroup$ actually assuming that the Dirichlet series converges for $\Re s >1$, it is enough to have only one-sided boundness $a_n \ge -C$ in the result above (note that $|a_n| \le C$ implies the convergence of the Dirichlet series in $\Re s >1$, but one-sided boundness doesn't necessarily, so we need this extra hypothesis - the issue though in this and @KConrad above is that one needs strong analytic information about the function on the whole boundary line, while the results of Littlewood-Hardy (and there are other sufficient conditions too) are just coefficient results, no analytic assumptions $\endgroup$
    – Conrad
    Jun 9, 2020 at 16:31
  • $\begingroup$ @Conrad the analytic information for the Fataou and Riesz theorems in my answer doed not need analytic information to the whole boundary (unit circle or ${\rm Re}(s) = 1$) but just to a specific point of interest. For example, consider the series $\sum z^n/n$, which has coefficients tending to $0$ and has an analytic continuation to each point on the unit circle except $z = 1$, since we know about the function $-\log(1-z)$. At each particular point on the unit circle other than $z = 1$, the Fatou theorem implies convergence of the series to $-\log(1-z)$. You don't need to know (contd.) $\endgroup$
    – KConrad
    Jun 10, 2020 at 1:15
  • $\begingroup$ analytic continuation of the function inside the unit disc to all of the unit circle minus $\{1\}$ to be able to say at a particular number $z$ on the unit circle, like $z = i$, that the series converges there to $-\log(1-z)$. By partial summation it's easy to see the series converges at each $z$ with $|z| = 1$ other than $z = 1$, but the partial summation does not make it evidence that the series is converging to $-\log(1-z)$. $\endgroup$
    – KConrad
    Jun 10, 2020 at 1:18
  • $\begingroup$ @KConrad - you are right and I stand corrected from the point of view of the full boundary, though the condition of analytic continuation even at a point is quite strong; Fatou's Theorem is indeed optimal as noted in your post since $a_n \to 0$ is obviously necessary while being sufficient under analytic continuation; the Hardy-Littlewood Tauberian theorems have stronger conditions on the coefficients (and just about applies to the $\log(1-z)$ example as $n|e^{int}/n| =1$) but require only radial convergence; same for the Dirichlet series and Riesz vs H-L, $\endgroup$
    – Conrad
    Jun 10, 2020 at 1:38
  • $\begingroup$ @Conrad what are a few Dirichlet series of practical interest to which the H-L Tauberian theorem applies and the Dirichlet series does not admit analytic continuation to a boundary point of interest? $\endgroup$
    – KConrad
    Jun 10, 2020 at 12:52

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