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.).

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.