TITLE: Do Abel summation and zeta summation always coincide?
QUESTION [25 upvotes]: This is a more focused version of Summation methods for divergent series.

Let $a_n$ be a sequence of real
  numbers such that $\lim_{x \to 1^{-}}
> \sum a_n x^n$ and $\lim_{s \to 0^{+}}
> \sum a_n n^{-s}$ both exist. (In
  particular, we assume that both of the
  sums in question converge in the
  appropriate region.) Need the limits
  be equal?


I've thought about this before, and am doing so again thanks to the linked question. Here are some things I've tried:
It is true if $a_n$ is periodic with average $0$.
If it is true for $a_n$, then it is true for the sequence $b_{kn+r}=a_n$, $b_m=0$ if $m \not \equiv r \mod k$.
It is true for $a_n=1$ if $n$ is an even square, $-1$ if $n$ is an odd square and $0$ otherwise. I tried to prove in general that, if it is true for $a_n$ then it is true for $b_{n^2}=a_n$, $b_m=0$ for $m$ not a square, but couldn't.
It appears to be true for $a_n = (-1)^n \log n$, although I didn't check all the details. 
I do not know any explicit sequence $a_n$ which obeys the hypotheses of the question and is not $(C, \alpha)$-summable for some $\alpha$. So it is possible that this is really a theorem about higher Cesaro summability. But I suspect such sequences do exist.
A natural generalization is: Let $a_n$, $\lambda_n$ and $\mu_n$ be three sequences of real numbers, with $\lambda_n$ and $\mu_n$ approaching $\infty$. If
$\lim_{s \to 0^{+}} \sum a_n e^{-\lambda_n s}$ and $\lim_{s \to 0^{+}} \sum a_n e^{-\mu_n s}$ both exist, need they be equal?
As far as I can tell, Wiener's generalized Tauberian theorem does not apply.

REPLY [8 votes]: G. H. Hardy, DIVERGENT SERIES, page 76...
"It is easy to give examples of series summable $(A,\log n)$ but not summable $(A)$:
we shall see, for example, that $\sum n^{-1-ci}$, where $c>0$, is such a series."
Before that, on page 73, we see if a series is both summable $(A,\log n)$ and summable $(A)$, then the two sums are equal.