Weak version of Karamata's Tauberian theorem 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.
 A: 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 $$\sum_{k = 0}^n a_k \asymp (1 - e^{-1/n})^{-\beta} \asymp n^\beta.$$
A: 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 .
