Q:

For bounded sequences, does convergence of the Abel means imply that for the Cesàro means?

See the title. This is true if the sequence is nonnegative; some Tauberian theorems which I was able to find give some more general sufficient conditions. I would like to know if this is true for arbitrary bounded sequences.
Recall that for a sequence $(a_n)$ with natural indices $n$, the Cesàro means are $\frac1{N}\sum\limits_{n=1}^N a_n$, and the Abel means are $(1-r)\sum\limits_n r^n a_n$.

A:

Yes. For the following modes of convergence 
you can  prove $(1)\Rightarrow (2)\Rightarrow (3)\Rightarrow (4)$.
$$\begin{eqnarray}
a_n &\to& a \qquad (1)\\ 
\sigma_n:={1\over n}\sum_{j=1}^{n} a_j &\to& a \qquad (2)\\
(1-r)\sum_{n=1}^\infty a_n r^n &\to& a\qquad (3)\\
(1-r)\sum_{n=1}^\infty \sigma_n r^n &\to& a\qquad (4)
\end{eqnarray}
$$
If $a_n$ is a bounded sequence then 
$(2)\Longleftrightarrow (3)$.
One direction is $(2)\Rightarrow (3)$ and the other follows
from Littlewood's Tauberian theorem ($(4)\Rightarrow (2)$) since $\sigma_{n+1}-\sigma_n$
is $O(1/n)$.

Reference. For further information and two  proofs of the "Abel to Cesàro" 
 theorem, see Chapter 1, sections 7, 11, and 12 of Tauberian Theory: A Century of Development 
by Jacob Korevaar. 

