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Let $(a_n)$ be a sequence of real numbers, and suppose that the real power series (function) $S(x):=\sum_{n=0}^{\infty} a_n x^n$ converges for every $x\in\mathbb{R}$.

$\mathbf{Question}$: Suppose that $\lim_{x\to \infty}S(x)$ exists finite. Is it possible to compute this limit in terms of the sequence $(a_n)$, i.e. in the form

\begin{equation} \lim_{n\to\infty} c_n \end{equation}

where $(c_n)$ is a suitable (and reasonable!) sequence build from $(a_n)$?

Any reference would be appreciated. Thanks in advance, Josh.

PS By reasonable I hope to exclude answers like: take the constant sequence defined starting from $(a_n)$ by the position $c_n:=\lim_{x\to \infty}\sum_{n=0}^{\infty} a_n x^n$.

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  • $\begingroup$ Do you want a process giving you the $c_n$ knowing the $a_n$ which works in every case? I very much doubt this is possible in such generality. $\endgroup$ – Loïc Teyssier May 15 '14 at 20:15
  • $\begingroup$ Hi Loïc. Of course before posting I've taught a little bit about the question. I convince myself that it is possible at least in many situations. And the key ingredients seems to me be the operations of convolution, Padé approximations and Cauchy integral formula. But my hope is still to find here someone that answer by saying "this is a well known result of the theory of..." and give me some reference :) $\endgroup$ – Josh May 15 '14 at 20:32
  • $\begingroup$ Maybe it would help if you could provide an example where you extract the limit from the $a_n$'s, and the power series can not easily be summed for arbitrary $x$. $\endgroup$ – Christian Remling May 15 '14 at 21:39
  • $\begingroup$ A meta-question: Can I answer to the question eventhough it is not an answer, but an approach that I would like to investigate with who will be interested? With a comment it is quite impossible to explain the idea... Thanks in advance. $\endgroup$ – Josh May 16 '14 at 16:03
  • $\begingroup$ Sure, I would like to hear about your approach. One option is also to write it in your question. $\endgroup$ – Joni Teräväinen May 16 '14 at 18:58
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There is no general way to construct such a sequence if we assume that $c_n$ depends only on $a_n,a_{n+1},...,a_{n+k}$ and $a_0,a_1,...,a_k$ for a fixed $k$ ($c_n$ must clearly depend at least on $a_0$). Indeed, if $S(x)$ has limit $L$ at infinity, then $S(x)+J_0(x^{k})-1$ has limit $L-1$ at infinity, where $J_0$ is the Bessel function of order $0$; $$J_0(x)=\sum_{m=0}^{\infty}(-1)^m \frac{x^{2m}}{4^m (m!)^2}.$$ However, the coefficients $a_0,a_1,...,a_k$ are the same for this new power series, and since $c_n\to L-1$ for the sequence corresponding to the new power series, also $c_{2nk-k-1}\to L-1$ but none of the numbers $c_{2nk-k-1}$ has changed (since they do not depend on those $a_m$ for which $2k$ divides $m$), which is a contradiction.

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  • $\begingroup$ Indeed,I expect that for every $n\in\mathbb{N}$, the $n$-element $c_n$ must be a function of $a_0,a_1,\ldots,a_n$. That's why I mentioned convolution. $\endgroup$ – Josh May 16 '14 at 15:38
  • $\begingroup$ In that case, the remainder formula for the Taylor series tells that one can take $c_n=a_0+a_1\log \log n+...+a_n(\log \log n)^n$ (for example), unless the derivatives of $S(x)$ grow too fast. I wouldn't expect a nice formula without any further assumptions. $\endgroup$ – Joni Teräväinen May 16 '14 at 18:56
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This is an approach that I found by "googling", "papering" and "wikipeding". When I realized the idea, I imagine myself as the first mathematicians looking at the way in which physicists made use of dirac distribution (at the very beginning).

What I mean, is that I'm still not able to justify any step of what follows, but according to many papers in Quantum Field Theory (a theory of which I do not know anything) it works in many many situations.

The idea is the following one (and believe me, to reach the following "cleaner" description was a terrible task).

Let us soppose that $\lim_{x\to\infty} S(x)=\ell\in \mathbb{R}-\{0\}$. This means that \begin{equation} \frac{1}{S(x)}=\frac{x}{x\sum_{n=0}^{\infty} a_n x^n}=\frac{x}{\sum_{n=1}^{\infty} b_n x^n}\to\frac{1}{\ell} \quad \text{when } x\to+\infty . \end{equation} Here we have defined $b_n:=a_{n-1}$ for every $n\in\mathbb{N}$, $n\geq 1$. In particular, for every $M\in\mathbb{N}$ one has \begin{equation} \frac{1}{S(x)}=\left(\frac{x^M}{\sum_{n=1}^{\infty} c(M)_n x^n}\right)^{1/M} \end{equation} where the sequence $(c(M))_n$ is given by the "M-times" polynomial convolution of the sequence $(b_n)$. Therefore \begin{equation} \frac{1}{S(x)}=\lim_{M\to\infty}\left(\frac{x^M}{\sum_{n=1}^{\infty} c(M)_n x^n}\right)^{1/M} . \end{equation}

Now starts the "strange" part. One writes \begin{equation} \frac{1}{S(x)}=\lim_{M\to\infty}\lim_{N\to\infty}\left(\frac{x^M}{\sum_{n=1}^{N} c(M)_n x^n}\right)^{1/M}. \end{equation} Next one choose $M=N$ to obtain \begin{equation} \frac{1}{S(x)}=\lim_{N\to\infty}\left(\frac{x^N}{\sum_{n=1}^{N} c(N)_n x^n}\right)^{1/N}. \end{equation} To finish, one pass to the limit for $x\to\infty$, and interchange the limits: \begin{equation} \lim_{x\to+\infty}\frac{1}{S(x)}=\lim_{N\to\infty}\left(\lim_{x\to+\infty}\frac{x^N}{\sum_{n=1}^{N} c(N)_n x^n}\right)^{1/N}=\lim_{N\to\infty} \frac{1}{c(N)_N^{1/N}} . \end{equation} Thus \begin{equation} \ell=\lim_{N\to\infty}c(N)_N^{1/N}. \end{equation}

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  • $\begingroup$ I'm suspicious about your $M=N$ step. In general, it is of course not true that $\lim_M\lim_N a_{MN} = \lim_N a_{NN}$. $\endgroup$ – Christian Remling May 17 '14 at 18:20
  • $\begingroup$ That's an interesting approach, but I am also suspicious about changing the limit. If one chooses $M=2N$, one ends up with $S(x)\to 0$. $\endgroup$ – Joni Teräväinen May 17 '14 at 19:03
  • $\begingroup$ Actually, I'm suspicious about everything :-) Of course there are some hypotheses which guarantee that it is possible to justify these two doubts, but the point is that these hypotheses must be imposed to the function $S(x)$ and not only on the coefficients $a_n$. By the way, maybe that one can avoid some of these steps in some way. Every further observation (maybe something that apparently could seems stupid) would be good. Who is interested can take a look to journals.aps.org/prl/abstract/10.1103/PhysRevLett.46.1255 $\endgroup$ – Josh May 17 '14 at 19:29
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If there is an uniform convergence, then, for example, $c_n=\sum_{k=0}^n a_kn^k$ will do. But no "reasonable" sequence will work in general, I agree with Loïc Teyssier on this. To say more, one has to know something about the series and the goal.

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