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Let $\sigma_n^{(k)}=\frac{1}{n+1}\sum_{j=0}^{n}\sigma_j^{(k-1)}$ and $\sigma_n^{(1)}=\frac{1}{n+1}\sum_{j=0}^{n}s_j.$

If $\lim_{n\to \infty}\sigma_n^{(k)}=L$ we call the sequence $(s_n)$ is summable $H_k$ to $L.$

Also the sequence $(s_n)$ is called Abel summable to $L$ if $\lim_{x\to1^{-}}(1-x)\sum_{j=0}^{n}s_jx^j=L.$

Does $H_k$ summability of $(s_n)$ to $L$ imply its Abel summability to $L?$

Also, are there any sequence which is Abel summable but not $H_k$ summable?

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The answer to your first question is yes, because Hölder summability (your method $H_k$) is equivalent to Cesàro summability, and Abel summability is stronger than all the Cesàro methods. These are theorems 49 and 55 in Hardy's Divergent Series (1949). So, if a sequence is $H_k$ summable, then it is $(C, k)$ summable and therefore Abel summable.

The answer to your second question is also yes. Hardy's example (p.109) defines $a_n$ by $$f(x) = e^{1/(1+x)} = \sum_{n \ge 0} a_n x^n$$ so that $f(x)$ exists for $|x| < 1$ and $f(x) \to e^{1/2}$ as $x \to 1$. But the sequence $a_n$ is not $O(n^k)$ for any $k$ since this would imply $f(x) = O((1 - |x|)^{-k-1})$ uniformly in the disk $|x| < 1$, whereas $f(x)$ tends to infinity like $e^{1/(1-|x|)}$ (which is much faster) when $x\to-1$ through real values. So, $a_n$ is not $(C, k)$ summable for any $k$, by the limitation theorem (46 op. cit.).

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  • $\begingroup$ I guess my first question is also true. If a sequence is $H_k$ summable then it is Abel summable. So $H_k$ summability of a sequence imply Abel summability. But if a sequence is Abel summable it may not $H_k$ summable, i.e. Abel summability does not imply $H_k$ summability. $\endgroup$ – Raio Dec 20 '15 at 9:52
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    $\begingroup$ I have already seen the book of Hardy. He gave the proof of the theorem "$(C,k)$ summability implies Abel summability" and he prove the two methods $(C,k)$ and $H_k$ are equivalent. $\endgroup$ – Raio Dec 20 '15 at 10:00
  • $\begingroup$ I wonder but can not directly prove the statement "$H_k$ summability implies Abel summability" without using equivalence of $H_k$ to $(C,k).$ $\endgroup$ – Raio Dec 20 '15 at 10:03
  • $\begingroup$ Oh! You're right, I got it backwards, let me fix that. Yes, if a sequence is $H_k$ summable, then it is $(C, k)$ summable and therefore Abel summable. $\endgroup$ – Unit Dec 20 '15 at 15:27
  • $\begingroup$ Yes, it is true, but i need a direct proof. $\endgroup$ – Raio Dec 20 '15 at 16:25

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