Divergent series which is Abel summable but not Euler summable

Question

It is said that:

Abel summation and Euler summation are not comparable.

We were able to find examples of divergent series which are Euler summable but not Abel summable, for instance $$ 1-2+4-8+16-\dots$$

However, we couldn't find any example of a divergent series which is Abel summable but not Euler summable.

Do you know such an example?

Thank you!

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EDIT: Dear Peter, this is the definition of Euler summation:

Let $\sum_{n=0}^\infty a_n$ be any series. The Euler transformation of this series is defined as: \begin{equation*} \sum_{n=0}^\infty \frac{1}{2^{n+1}}b_n\quad\text{ with }\quad b_n:=\sum_{k=0}^n\left(\begin{array}[h]{c} n \\ k \end{array}\right) a_k \end{equation*}

The series $\sum_{n=0}^\infty a_n$ is called Euler summable if the Euler transformation of this series $$\sum_{n=0}^\infty \frac{1}{2^{n+1}}b_n$$ is converges in the usual sense.

The Euler sum is then given by $$\sum_{n=0}^\infty \frac{1}{2^{n+1}}b_n.$$

Answer 1

From the Wikipedia article,

Euler summation is essentially an explicit form of analytic continuation. If a power series converges for small complex $z$ and can be analytically continued to the open disk with diameter from $-1/2$ to $1$ and is continuous at $1$, then its value at $1$ is called the Euler sum of the series $a_0 + a_1 + \ldots$.

Whereas Abel summation consists of taking the limit of $ f(z)\equiv \sum_{n=0}^{\infty} a_n z^n $ as $z$ approaches $1$ from below along the real axis. For a series to be Abel-summable but not Euler-summable, it has to be that $f(z)$ has a limit as $z\rightarrow 1^-$ along the real axis, but $f(z)$ is not continuous at $z=1$. An example would be $f(z)=\exp\left(-\frac{z}{1-z}\right)$, which has an essential singularity at $z=1$, but for which $\lim_{z\rightarrow 1^-}$ exists and is equal to $0$.