# 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 $(1-r)\sum\limits_n r^n a_n$.

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 @J.M.: you removed the set over which the sum is taken in the definition of the Cesaro means. Well, let it be, let us hope that the readers know this. – urosIV Nov 22 '11 at 17:55 I'll restore it. I was going for consistency since it looked fine for you to omit the index terms in the Abel mean... – J. M. Nov 22 '11 at 17:59 You are welcome to make it looking perfect)) – urosIV Nov 22 '11 at 18:00 In any event: I imagine Hardy's book on divergent series ought to have results on this. Have you checked that already? – J. M. Nov 22 '11 at 18:02 Yes I did. I did not find this result and they don't seem to pay special attention to bounded sequences. – urosIV Nov 22 '11 at 18:05

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 $O(1/n)$.