# 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?

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.

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David, in the statement of the problem, did you mean to have a sum in the second limit? (I'm pretty sure you are looking at a Dirichlet series, but I don't have editing power.) – Pace Nielsen Mar 26 '10 at 17:05
@PN: I'm sure that he did, and so I've gone ahead and added the $\sum$. – Theo Johnson-Freyd Mar 26 '10 at 17:20
That indeed is what I meant. Thanks Theo! – David Speyer Mar 26 '10 at 21:26

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.

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Possibly dumb question: are you sure we can interchange summation and integration? The obvious fact is that $\sum a_n/n^s = (1/\Gamma(s)) \sum \int \cdots$, while your argument needs $\int \sum$. – David Speyer Apr 28 '10 at 12:23
You're right; it's a nontrivial point. I added an argument to the post justifying it. – Brad Rodgers Apr 28 '10 at 19:11
Actually, I had another think over this today, and I'm making a mistake in using dominated convergence. If nothing else, one (I think) can get by establishing the interchange of summation and integration for large enough $s$, and then invoke analytic continuation, but the result ought to have a real variable proof. Perhaps until I've thought about it more thoroughly, Gerald's reference to Hardy is the place to go... – Brad Rodgers Apr 29 '10 at 19:29
Alright. I think I fixed everything. I did finally take a look at Hardy; I'm not sure if his proof is properly 'homotopic' to this one, but the main idea is still to take Mellin transforms. – Brad Rodgers Apr 30 '10 at 19:58

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.

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Your last sentence has a missing «don't», I guess! – Mariano Suárez-Alvarez Apr 28 '10 at 19:14

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 $s = 1$.

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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.

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Thanks! Between you and Brad Rogers, this seems to be thoroughly answered. – David Speyer Apr 29 '10 at 10:46