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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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up vote 3 down vote accepted

I guess this question is from "Multiplicative Number Theory" by Montgomery. As I studied this book too, I was stuck with this problem. Finally, after 6 years, I have an answer to this.

A good reference to my answer is "The General Theory of Dirichlet Series" by Hardy and Riesz. Chapter 4.

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.

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(+1) Now the answer is clear, thanks! – Sangchul Lee Mar 17 '13 at 4:10
You're welcome! Surprised by real fast reply. – i707107 Mar 17 '13 at 4:12

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