An example of a series that is Abel-summable but not Cesàro-summable

Question

In the book of Hardy, Theorem 56 (page 108), it is stated that

There are series that are Abel-summable but not $(C, k)$-summable for any $k$.

We are looking for an example of a series that is Abel-summable but not higher Cesàro-summable. We could not find any example ourselves, as well as in any book. We found only examples concerning Abel summation and simple Cesàro summation, but not higher Cesàro summation.

Thank you!

Answer 1

$\def\e{\mathrm{e}}$Such example is right presented on page 109: Define $\{a_n\}$ by$$ f(z) = \exp\left( \frac{1}{1 + z} \right) = \sum_{n = 0}^∞ a_n z^n. \quad \forall |z| < 1 $$ Since $z = 1$ is regular point of $f$, then$$ \lim_{\substack{z → 1\\|z| < 1}} \sum_{n = 0}^∞ a_n z^n = \lim_{\substack{z → 1\\|z| < 1}} f(z) = \sqrt{\e}. $$ To prove that $\sum\limits_{n = 0}^∞ a_n$ is not (C, $k$)-summable for any $k > 0$, it suffices to prove that $a_n ≠ O(n^k)$ $(n → ∞)$. Suppose $a_n = O(n^k)$ $(n → ∞)$, then$$ \left| \exp\left( \frac{1}{1 + z} \right) \right| = |f(z)| \leqslant \sum_{n = 0}^∞ a_n |z|^n \leqslant C_1 \sum_{n = 0}^∞ n^k |z|^n \leqslant \frac{C_2}{(1 - |z|)^{k + 1}}, \quad \forall |z| < 1 $$ which leads to contradiction by making $z \in \mathbb{R}$, $z → -1 + 0$.