If a series has a well-defined Cesaro sum, then it has a well-defined Abel sum and they are equal. I think I first learned this from Hardy's Divergent Series; the proof is short enough to give here.
Let $a_i$ be the series in question, let $s_m = sum_{i=0}^m a_i$ and $c_n = sum_{m=0}^n s_m$. The claim that the Cesaro sum is well defined is that
$$c_n = (n+1)(L + o(1)).$$
Let $A(x) = sum a_i x^i$. Then $sum c_i x^i = A(x)/(1-x)^2$ so
$$ A(x) = (1-x)^2 sum_{n=0}^{infty} (L (n+1) + o(n+1)) x^n$$
where the $o$ is as $n to infty$, independent of $x$.
But
$$sum_{n=0}^{infty} (n+1) x^n = 1/(1-x)^2$$
so
$$A(x) = L (1-x)^2/(1-x)^2 + o left( (1-x)^2/(1-x)^2right) = L+ o(1) quad mbox{as} x to 1^{-}.$$
So the Abel sum of $a_i$ is also $L$.
Deleted an argument that Cesaro summability implies zeta summability; not sure I can sum by parts where needed.
I want to say that I feel guilty writing up special cases like this.
I have a vague impression that there is a very general philosophy here, something like Wiener's generalized Tauberian theorem. (But presumably easier, since we are generalizing Abel's theorem, not Tauber's.)
I'm hoping that someone will come by and write up an exposition of it.
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