An apparently simple question (behaviour at infinity of a power series)

Question

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 $c_n:=\lim_{x\to \infty}\sum_{n=0}^{\infty} a_n x^n$.

Answer 1

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.

Answer 2

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}

Answer 3

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.