# Is there a superpolynomial sequence which is Abel-summable?

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.

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 $$O(t^{-c-1})$$.