This is a more focused version of [Summation methods for divergent series](https://mathoverflow.net/questions/19201/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](http://en.wikipedia.org/wiki/Cesaro_summation) 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.



Do Abel summation and zeta summation always coincide?
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.


