# Abelian theorem regarding Riesz summability

This is my first time to post something here. If there is anything wrong, please inform me... Anyway, here is my question:

Let $k$ be a nonnegative integer. We say a sequence $(a_n)$ is $(R, k)$-summable to $a$ if

$\displaystyle \lim_{x\to\infty} \sum_{n \leq x} a_n \left( 1 - \frac{\log n}{\log x} \right)^k = a,$

and we denote $\sum a_n = a \ (R, k)$.

It is easy to show that $(R, 0)$-summability is equivalent to the ordinary summability, and $\sum a_n = a \ (R, k)$ implies $\sum a_n = a \ (R, j)$ for all $j \geq k$.

My question here is like this: Let $\alpha (s) = \sum a_n n^{-s}$. If $\sum a_n = a \ (R, k)$, then does the limit $\lim_{s \to 0^+} \alpha (s)$ exist? If so, then does the limit coincide with $a$?

I was able to prove that $\alpha (s)$ can by analytically continued for $\Re (s) > 0$, and for this continuation, we have $\alpha (s) \to a$ as $s \to 0^{+}$. But it is not immediate, or even not sure if this guarantees the convergence of $\sum a_n n^{-s}$ for $\Re (s) > 0$. Or is there any other way to proof of disprove that $\sum a_n n^{-s}$ exists for $\Re (s) > 0$?

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The answer for the first question is NO, because the existence of the limit as $s\rightarrow 0$ is not guaranteed. Consider the following, $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s}.$$ This series is known to be Cesaro summable of order $k$ if $-k<\sigma=\textrm{Re}(s)$. (C,k) summability implies (R,k) summability. Then as you can see, the Dirichlet series has abscissa of convergence $0$. Hence, the conclusion cannot be true, use Cesaro summability of order 2 at $s=-1$.
As you said already, (R,k) summability implies the existence of analytic continuation to $\textrm{Re}(s)>0$, and the limit equals the sum.