I first posted this on mathematics. However, I got no answer there and it seems adapted here too. Also, it seems to be harder than I first thought.

Karamata's Tauberian theorem states the following. Let $A(z)=\sum a_nz^n$ be a power series with non-negative coefficients $a_n$ and radius of convergence 1. Let $\beta>0$. Then, as the real variable $s\in [0,1]$ tends to 1, $\sum_{n\geq 0}a_ns^n\underset{s\to 1}{\sim} c/(1-s)^\beta$ if and only if $\sum_{k=0}^na_k\underset{n\to \infty}{\sim}c'n^{\beta}$, where $c$ and $c'$ are determine each other. Moreover, if $a_n$ is non-increasing, then one can replace $\sum_{k=0}^na_k$ with $a_n$, that is, we have $a_n\sim c''n^{\beta-1}$. See for example Corollary 1.7.3 in Bingham, Goldie and Teugel's book Regular variation.

I'm wondering if the following is true. For two functions $f$ and $g$, write $f\asymp g$ if there exists $C\geq 0$ such that $f\leq Cg$ and $g\leq Cf$. Let $A(z)=\sum a_nz^n$ be a power series with non-negative coefficients $a_n$ and radius of convergence 1. Then, $\sum_{n\geq 0}a_ns^n \asymp 1/(1-s)^\beta$ for $s\in [0,1]$ if and only if $\sum_{k=0}^na_k\asymp n^{\beta}$ if and only if $a_n\asymp n^{\beta-1}$. The implicit constants are asked not to depend on $s$ and $n$ respectively.

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    $\begingroup$ Surely there are typos in the statements of Karamata's theorem: you mean $A(s) \sim c/(1-s)^{\beta}$ (twice)? $\endgroup$ Mar 1, 2020 at 23:53
  • $\begingroup$ @DavidHandelman Sure, thanks for noticing ! I edited accordingly $\endgroup$
    – M. Dus
    Mar 2, 2020 at 8:52
  • $\begingroup$ The asymptotics $a_n \approx n^{\beta}$ and $\sum_{k=0}^n a_k \approx n^{\beta}$ sound inconsistent; $a_n$ should behave like $n^{\beta-1}$. Also, if $(a_n)$ is decreasing, then $\beta$ can be at most 1. $\endgroup$ Mar 2, 2020 at 9:29
  • $\begingroup$ I do not have access to Bingham–Goldie–Teugels book right now, but did you check in the chapter on "$O$-regular variation" (Chapter 2, I believe)? At least in the continuous case (the Laplace transform instead of the power series), both Karamata's Tauberian theorem and monotone density theorem have their couterparts for the "$O$-regular variation" (that is, roughly speaking, with "$\sim$" replaced by "$\asymp$"). $\endgroup$ Mar 2, 2020 at 20:07
  • $\begingroup$ @GiorgioMetafune Of course you're right, thank you very much. I'm particuarly interested in the case where $\beta <1$. $\endgroup$
    – M. Dus
    Mar 3, 2020 at 15:10

2 Answers 2


This seems to follow easily from de Haan–Stadtmüller Theorem; see Theorem 2.10.2 in the Bingham–Goldie-Teugels book:

Theorem: Let $U$ be non-decreasing, and vanish on $(-\infty, 0)$. The following are equivalent:

(i) $U \in OR$;

(ii) $\hat{U}(1/\cdot) \in OR$;

(iii) $\hat{U}(1/t) \asymp U(t)$ ($t \to \infty$).

Here $\hat{U}(\lambda) = \int_{[0, \infty)} e^{-\lambda x} U(dx)$ is the Laplace–Stieltjes transform of the measure $U(dx)$, and $OR$ stands for the class of $O$-regularly varying functions at infinity, that is, positive functions $f$ such that $$\limsup_{x \to \infty} \frac{f(\lambda x)}{f(x)} < \infty$$ for every $\lambda \in (0, \infty)$.

Now suppose that $$A(z) = \sum_{n = 0}^\infty a_n z^n $$ with $a_n \geqslant 0$. If we let $U(x) = \sum_{k = 0}^n a_k$ for $x \in [n, n+1)$, then $$ \hat{U}(\lambda) = \sum_{n = 0}^\infty a_n e^{-\lambda n} = A(e^{-\lambda}) . $$ Thus the above theorem reads as follows:

Corollary: With the above notation, the following are equivalent:

(i) $\sum_{k = 0}^{\lfloor \cdot \rfloor} a_k \in OR$;

(ii) $A(e^{-1/\cdot}) \in OR$;

(iii) $A(e^{-1/t}) \asymp \sum_{k = 0}^{\lfloor t \rfloor} a_k$ ($t \to \infty$).

Clearly, if $$\sum_{k = 0}^n a_k \asymp n^\beta,$$ then (i) holds, and by (iii) we have $$A(e^{-1/t}) \asymp t^\beta.$$ On the other hand, if $$A(s) \asymp (1 - s)^{-\beta}$$ as $s \to 1^-$, then (ii) is satisfied, and by (iii) we have $$\sum_{k = 0}^n a_k \asymp (1 - e^{-1/n})^{-\beta} \asymp n^\beta.$$

  • $\begingroup$ Well thank you very much. I was not aware of this result and I should have looked more carefully at Bingham–Goldie-Teugels book first. $\endgroup$
    – M. Dus
    Mar 3, 2020 at 15:19
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    $\begingroup$ You are welcome. The Regular variation book is very useful, but I always have problems finding the result I need there. $\endgroup$ Mar 3, 2020 at 20:18

This is too long for a comment but I did not check all the details. I think that the solution can be found following the arguments in the book of Titchmarsh "The theory of functions" pp. 224 and following. First of all, the series expansion of $(1-x)^{-\beta}$ has positive coefficients $b_n \approx n^{\beta-1}$. Therefore, if $a_n \approx n^{\beta-1}$, then $F(x) \approx (1-x)^{-\beta}$. More generally, the same holds if $s_n:=\sum_{k=0}^n a_k \approx n^\beta$, using $F(x)=(1-x)\sum_{n=1}^\infty s_n x^{n-1}$. The converse, that is going from $F$ to $(s_n)$, should follow from the preceeding consideration by adapting the explanations in 7.52, where the author shows how to obatin weaker forms of Karamata's theorem with $\beta=1$, using more elementary tools .

  • $\begingroup$ Nice answer. Thank you very much. $\endgroup$
    – M. Dus
    Mar 3, 2020 at 15:21

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