I first posted this on [mathematics](https://math.stackexchange.com/questions/3555306/weak-tauberian-theorem). 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](https://www.cambridge.org/core/books/regular-variation/A71D6C4A65CFC1806E1DB7109128A358).


**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.



Weak version of Karamata's Tauberian theorem
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 $$\sum\_{k = 0}^n a\_k \asymp (1 - e^{-1/n})^{-\beta} \asymp n^\beta.$$


