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?
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.).
