Do Abel summation and zeta summation always coincide? 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.
 A: 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.
A: I don't have anything to say on this Abel theorem business for power series and Dirichlet series off the top of my head, but I want to point out that there is no Abel theorem for infinite products.
In 1908, Hardy (see Collected Papers Vol. VI pp. 263--271) gave an example of a 
sequence of numbers $a_n$ such that $\prod_{n \geq 1} (1+a_n)$ converges and 
$$
\lim_{x \rightarrow 1^{-}} \prod_{n \geq 1} (1+a_nx^n) = 2\prod_{n \geq 1}(1+a_n).
$$
Note the extra factor of 2 on the right.  On the Euler product side, for ${\rm Re}(s) > 1$ consider the function 
$$
\frac{\zeta(ms-(m-1))}{\zeta(s)} = \prod_{p} \frac{1-p^{-s}}{1 - p^{m-1-ms}}, 
$$
where $m \geq 1$. While the right side at $s = 1$ is equal to 1, using the analytic expression on the left side we can check that the limit of this function as $s \rightarrow 1^{+}$ is $1/m$. Let $m$ be a positive integer that is at least 2 to get a discontinuous Euler product at $s = 1$.
A: I'm very dubious, but I don't have handy a copy of G.H. Hardy, Divergent Series, 1949 — it's my primary reference for this type of question.
As you've correctly pointed out, the generalized question is whether $$\lim_{s\to 0^+} \sum a_n e^{-\lambda_n s} = \lim_{s\to 0^+} \sum a_n e^{-\mu_n s}$$ for $\lambda_n,\mu_n$ both increasing to $\infty$.  And for this the answer is a resounding "not always".  Consider for example $e^{-\lambda_n s} = x^n$ and $e^{-\mu_n s} = x^{\lfloor 3n/2 \rfloor}$, for some changes of variables $x(s)$, and $a_n = (-1)^n$.  Then the LHS is:
$$ 1 - x + x^2 - x^3 + x^4 - \dots = \frac 1 {1+x} \to \frac12 $$
whereas the RHS is:
$$ 1 - x + x^3 - x^4 + x^6 - \dots = \frac 1 {1 + x + x^2} \to \frac13 $$
In general, by "spacing" the alternating sequence $(-1)^n$ correctly, you can get its Abel-style summations to converge to any number in $[0,1]$.  This is a more continuous version of "divergent series are not associative", just like "conditionally convergent series are not commutative".  The point is that by twiddling the coefficients in $\lim \sum a_n e^{-\lambda_n s}$, you effectively twiddle the associatization.  But it might just happen that the $x^n$ and the $n^{-s}$ spacings are both sufficiently regular that they give the same answer.
Hardy [op. cit.] does provide a number of theorems about when these different summation methods agree, although his main focus in the book is when these different summation methods give the same answer.
A: I think the answer is 'yes.' I don't have a suitably general reason why this is the case, although surely one exists and is in the literature somewhere.
At any rate, for the problem at hand, we have for $s > 0$
$$\sum \frac{a_n}{n^s} = \frac{1}{\Gamma(s)}\int_0^\infty \sum a_n e^{-nt} t^{s-1} dt.$$
Edit: the interchange of limit and sum used here requires justification, and this is done below. Supposing that $\sum a_n x^n \rightarrow \sigma$, we may write $\sum a_n e^{-nt} = (\sigma + \epsilon(t))\cdot e^{-t}$ where $\epsilon(t) \rightarrow 0$ as $t \rightarrow 0$, and $\epsilon(t)$ is bounded for all t. In this case
$$\sum \frac{a_n}{n^s} = \sigma + O\left(s\int_0^\infty \epsilon(t) e^{-t} t^{s-1} dt\right)$$
Showing that error term tends to $0$ is just a matter of epsilontics; for any $\epsilon > 0$, there is $\Delta$ so that $|\epsilon(t)| < \epsilon$ for $t < \Delta$. Hence
$$\left|s\int_0^\infty \epsilon(t) e^{-t} t^{s-1} dt \right| < s\epsilon \int_0^\Delta t^{s-1} dt + s\int_\Delta^\infty e^{-t}t^{s-1}dt < \epsilon \Delta^s + s\int_\Delta^\infty e^{-t} t^{-1} dt.$$
Letting $s \rightarrow 0$, our error term is bounded by $\epsilon$, but $\epsilon$ of course is arbitrary.
Edit: Justifying the interchange of limit and sum above is surprisingly difficult. We will require 
Lemma: If for fixed $\epsilon > 0$, the partial sums $D_{\epsilon}(N) = \sum_{n=1}^N a_n/n^\epsilon = O(1),$ then
(a) $A(N) = \sum_{n \leq N} a_n = O(n^\epsilon)$, and
(b) $\sum_{n \leq N} a_n e^{-nt} = O(t^{-\epsilon})$,
where the O-constants depend on $\epsilon.$
This, with the hypothesis that $\sum a_n/n^s$ converges for all $s > 0$, imply the conclusions a) and b) for all positive $\epsilon$.
To prove part a), note that 
$$\sum_{n \leq N} a_n = \sum_{n \leq N} a_n n^{-\epsilon}n^\epsilon = \sum_{n \leq N-1} D_{\epsilon}(n) (n^\epsilon - (n+1)^\epsilon) + D_\epsilon(N)N^\epsilon,$$
which is seen to be $O(N^\epsilon)$ upon taking absolute values inside the sum.
To prove part b), note that 
$$t^\epsilon \sum_{n \leq N} a_n e^{-nt} = t^\epsilon \sum_{n=1}^{N-1} A(n)(e^{-nt} - e^{-(n+1)t}) + t^\epsilon A(N) e^{-Nt} = O \left( \sum_{n\leq N} (tn)^\epsilon e^{-nt}(1-e^{-t}) + (tN)^\epsilon e^{-Nt}\right).$$
Now, $(tN)^\epsilon e^{-Nt} = O(1)$, and 
$$\sum_{n\leq N} (tn)^\epsilon e^{-nt}(1-e^{-t}) = 2^\epsilon(1-e^{-t}) \sum_{n\leq N} (tn/2)^\epsilon e^{-nt/2} e^{-nt/2} = O\left(\frac{1-e^{-t}}{1-e^{-t/2}}\right) = O\left(\frac{1}{1+e^{t/2}}\right) = O(1),$$
and this proves b).
We use this to justify interchanging sum and integral as follows: note that
$$\sum_{n=1}^N \frac{a_n}{n^s} = \frac{1}{\Gamma(s)}\int_0^\infty \sum_{n=1}^N a_n e^{-nt} t^{s-1} dt,$$
and therefore
$$\frac{1}{\Gamma(s)}\int_0^\infty \lim_{N\rightarrow\infty}\sum_{n=1}^N a_n e^{-nt} t^{s-1} dt =  \frac{1}{\Gamma(s)}\int_0^1 \lim_{N\rightarrow\infty}\sum_{n=1}^N a_n e^{-nt} t^{s-1} dt + \frac{1}{\Gamma(s)}\int_1^\infty \lim_{N\rightarrow\infty}\sum_{n=1}^N a_n e^{-nt} t^{s-1} dt.$$
In the first integral, note that for $\epsilon < s$, $\sum_{n \leq N} a_n e^{-nt} t^{s-1} = O(t^{s-\epsilon -1})$ for all $N$.  So by dominated convergence in the first integral, and uniform convergence of $e^t \sum_{n=1}^N a_n e^{-nt}$ for $t \geq 1$ in the second, this is limit is
$$\lim_{N\rightarrow\infty}\frac{1}{\Gamma(s)}\int_0^1 \sum_{n=1}^N a_n e^{-nt} t^{s-1} dt + \lim_{N\rightarrow\infty}\frac{1}{\Gamma(s)}\int_1^\infty \sum_{n=1}^N a_n e^{-nt} t^{s-1} dt = \lim_{N\rightarrow\infty} \sum_{n=1}^N a_n \frac{1}{\Gamma(s)}\int_0^\infty e^{-nt}t^{s-1} dt.$$ 
This is just $\sum_{n=1}^\infty \frac{a_n}{n^s}$.
Note then that we do not need to assume from the start that the infinite Dirichlet sum tends to anything as $s \rightarrow 0$; once it converges for each fixed $s$, that is implied by the behavior of the power series.
