Limit at Infinity of Maclaurin Series

Let $f(x) = \sum_{n=1}^\infty a_n x^n$. What is $\lim_{x\rightarrow \infty} f(x)$ in terms of the $a_i$?
That question may be too broad, so here are some restrictions: Assume f(x) is continuous (and therefore well-defined) on all of $\mathbb{R}^+$ and the limit exists in the extended reals.
For example, if I were given the sequence $$\{a_i\}_{i=0}^\infty = \{1,-1,1/2,-1/6,1/24,\cdots, (-1)^i / i!, \cdots \}$$, I would have no idea what its limit at infinity is, but as soon I knew it was $e^{-x}$, taking the limit would be easy ($\lim_{x \rightarrow \infty} e^{-x} = 0$). The same is true for $e^x$.

Any help is appreciated.
Thanks

edit: To make the question simpler, I would be happy with determining if the limit is finite or infinite.

• All you have to do is change $a_{1000000}$ by $0.0000000000001$ to change the limit from zero to "does not exist", so I don't expect you'll get any useful answer to your question. – Gerry Myerson Feb 6 '18 at 1:42
• "Assume f(x) is continuous (and therefore well-defined) on all of $\mathbb{R}^+$" Redundant. If that series converges for all $x>0,$ then $f$ is continuous, $C^\infty,$ analytic, whatever you throw at it. – zhw. Feb 6 '18 at 1:48
• @zhw. Of course you're correct. But the OP is asking how to ascertain the limit as $x\to\infty$ from the series representation of a function. – Mark Viola Feb 6 '18 at 1:52
• The function is entire on $\mathbb{C}$ and, unless it is a constant, it has an essential singularity at complex infinity. That means that the real limit as $x \to \infty$, if it exists, is approaching the singularity from just one direction. – 6005 Feb 6 '18 at 4:28
• Cauchy Hadamard theorem implies that $\lim_{n \to \infty} |a_n|^{1/n} \to 0$, so one can try to write $a_n = \pm (c_n)^n$, where $c_n \to 0$. – 6005 Feb 6 '18 at 4:31

We can write:

$$\lim_{x\to\infty}f(x) = f(0) + \int_0^{\infty}f'(x)dx\tag{1}$$

We can express the integral in terms of the analytically continued series expansion coefficients using Ramanujan's master theorem. A heuristic derivation can be given as follows. A series expansion of a function $g(x)$:

$$g(x) = \sum_{k=0}^{\infty}(-1)^k\frac{c_k}{k!} x^k$$

can be formally written as:

$$g(x) = \sum_{k=0}^{\infty}(-1)^k\frac{x^k}{k!} E^{k}c_0 = \exp\left(-E x\right)c_0$$

where the operator $E$ acts on the coefficients as follows:

$$E c_{k} = c_{k+1}$$

This expression allows us to formally express integrals of functions in terms of their series expansion coefficients. E.g., we have:

$$\int_{0}^{\infty}x^sg(x)dx = \int_{0}^{\infty}x^s\exp\left(-E x\right)c_0dx = s! E^{-(s+1)}c_{0} = s! c_{-s-1}$$

The integral on the r.h.s. of (1) is thus $c_{-1}$, where $c_k$ is the coefficient of $(-1)^k\frac{x^k}{k!}$ of $f'(x)$, this is given by $-f(0)$. The limit in (1) is thus equal to zero.

This looks like an nonsensical result, because how can the limit always be zero, as Masacroso also points out in the comments. But we need to take into account that this is a purely formal result. A limit of an analytic function to infinity in the complex plane can always be taken in such a way that it yields zero, because of the essential singularity at infinity, so there is no real problem here. This means that the integral in (1) should in general be taken along a suitable contour to infinity.

• $\frac{\sin(x)}{x}$? – W. mu Feb 6 '18 at 2:28
• @W.mu I corrected some errors, it turns out that the limit will always be zero, at least in this formal sense. – Count Iblis Feb 6 '18 at 2:48
• but this would imply that any analytic function have by limit at infinity zero, what is not true. – Masacroso Feb 6 '18 at 3:02
• @Masacroso I've added an explanation. – Count Iblis Feb 6 '18 at 19:56
• @CountIblis but the question is "what is $\lim_{x\to\infty}f(x)$ in terms of $a_i$?". I dont know how a formalization that makes a "limit" zero is related to the question. – Masacroso Feb 6 '18 at 20:11