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A sequence $a_n \in \mathbb{C}, \ n = 1, 2, 3, \dots$ is Abel-summable if for all $|x| < 1$ the sum $$g(x) = \sum_{n = 1}^{\infty} a_n x^n$$ converges and the limit $\lim_{x \to 1^{-}} g(x)$ exists. In the case the limit is called the Abel sum of the sequence $a_n$. Notice that for $g(x)$ to converge, $a_n$ must grow subexponentially.

Question: Is every Abel-summable sequence of polynomial growth, that is there exists $C_1, C_2 > 0$ such that for all $n$ we have $| a_n | \leq C_1 n^{C_2}$?


Here is the motivation for the question: say that $a_n$ is zeta-regularizable if for $s$ with $\mathrm{Re} (s)$ large enough the Dirichlet series $$f(s) = \sum_{n = 1}^{\infty} \frac{a_n}{n^s}$$ converges, $f(s)$ has an analytic continuation to $\mathrm{Re} (s) > 0$, and the limit $\lim_{s \to 0^{+}} f(s)$ exists. In this case, the limit is called the zeta-regularized sum of the sequence $a_n$. Notice that for $f(s)$ to converge for some $s$ we need $a_n$ to be of polynomial growth, that is $$a_n = \mathcal{O} (n^c)$$ for some $c > 0$. I recently learned of the interesting fact that if $a_n$ is of polynomial growth and is Abel-summable, then it is also zeta-regularizable, and its zeta-regularized sum is the same as its Abel sum (the basic idea is that the Mellin transform of $g(e^{- x})$ is equal to $\Gamma (s) f(s)$).

From this, we can see that these two summation methods are consistent, that is whenever they both assign a finite value to two series these values are equal. Notice now that an equivalent phrasing of the question is as follows: is every Abel summable sequence also zeta-regularizable? That is, is zeta-regularization strictly stronger than Abel summation?

My gut instinct says no, but I've tried a bit and I couldn't find an example.

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Let $$f(x):=\sum a_nx^n=\exp\left(\frac1{1+x}\right), |x|<1.$$ Then $a_n$ is Abel summable to $\sqrt{e}$, but if we had $a_n=O(n^c)$, the value of $f(-1+t)$ for small $t$ would be bounded by $O(t^{-c-1})$.

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