Let $$ f(x) = \sum_{n=0}^{\infty} a_n x^n $$ and suppose that the radius of convergence of this series is infinite. Is there a general method to know whether $\lim_{x \rightarrow \infty} f(x)$ exists and to calculate the limit if it exists. For example, we know that $$ \lim_{x \rightarrow \infty} \sum_{n=0} \frac{(1)^n x^n}{n!} = 0 $$ since the LHS is equal to $e^{x}$. However, most infinite series cannot be represented in terms of standard functions, special functions, or solution of a differential equation, so I wonder if there is any way to study $\lim_{x \rightarrow \infty} f(x)$ directly from the coefficients $ a_n$.

6$\begingroup$ I would be surprised if the large$x$ limit could be obtained directly from the coefficients, think about comparing $\sum_n (1)^n x^{2n}/(2n!)$ and $\sum_n (1)^n x^{2n1}/(2n!)$  very similar sums, both with infinite radius of convergence, one tends to zero while the other has no large$x$ limit at all. $\endgroup$– Carlo BeenakkerJun 1, 2021 at 15:17

5$\begingroup$ There is no reasonable method. Functions like $e^{x}$ and $\sin x$ are very rare exceptions. Most power series are unbounded when $x\to+\infty$. $\endgroup$– Alexandre EremenkoJun 1, 2021 at 15:52

$\begingroup$ For some series $\sum_n a(n) x^n = \sum_n e^{n \log x + \log a(n)}$, the coefficients $a(n)$ may be nice enough that you could do a saddlepoint type approximation: find the local maxima of $n \log x + \log a(n)$ as a function of $n$ (for fixed $x$) and hope that the values of $n$ far away from these maxima will only contribute negligibly to the sum. This sometimes helps find the growth rate of the series. If you're lucky, you might find that the growth remains bounded. $\endgroup$– Igor KhavkineJun 1, 2021 at 22:17
1 Answer
I suggest looking into Ramanujan's Master theorem https://en.wikipedia.org/wiki/Ramanujan%27s_master_theorem
It's not exactly what you're looking for but there is a restricted case where you can say the limit $\sum_n a(n)x^n \to 0$ as $x \to \infty$.
Essentially it is, if $F(z) : \mathbb{C}_{\Re(z) > 1} \to \mathbb{C}$ is holomorphic, and $F(z) \le C e^{\rho \Re(z) + \tau\Im(z)}$ for some $C,\rho,\tau$ and $\tau < \pi/2$; then the series,
$$ f(x) = \sum_n F(n)(1)^n\frac{x^n}{n!}\\ $$
satisfies,
$$ \int_0^\infty f(x) x^{z1} \,dx = \Gamma(z) F(z)\\ $$
And in particular, $f(x) \to 0$ as $x \to \infty$. You can also be assured that,
$$ f(w) \to 0\,\,\text{as}\,\,w\to\infty\,\,\text{for}\,\,\arg(w) < \pi/2  \tau\\ $$
Where Ramanujan's theorem will hold as an integration on this sector.