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.

  • $\begingroup$ 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.) $\endgroup$ Mar 26, 2010 at 17:05
  • 1
    $\begingroup$ @PN: I'm sure that he did, and so I've gone ahead and added the $\sum$. $\endgroup$ Mar 26, 2010 at 17:20
  • $\begingroup$ That indeed is what I meant. Thanks Theo! $\endgroup$ Mar 26, 2010 at 21:26

4 Answers 4


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.

  • $\begingroup$ 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 $\int \sum$. $\endgroup$ Apr 28, 2010 at 12:23
  • $\begingroup$ You're right; it's a nontrivial point. I added an argument to the post justifying it. $\endgroup$ Apr 28, 2010 at 19:11
  • $\begingroup$ 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... $\endgroup$ Apr 29, 2010 at 19:29
  • $\begingroup$ 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. $\endgroup$ Apr 30, 2010 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.

  • 2
    $\begingroup$ Your last sentence has a missing «don't», I guess! $\endgroup$ Apr 28, 2010 at 19:14

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.

  • $\begingroup$ Thanks! Between you and Brad Rogers, this seems to be thoroughly answered. $\endgroup$ Apr 29, 2010 at 10:46

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 $s = 1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.