Question: Given a function $f: \mathbb{R} \to \mathbb{R}$ such that the MacLaurin series exists and equals the function for every $x \in \mathbb{R}$, and such that for all $n \ge n_0$, $n_0$ some positive integer, the $n$th term of the series does not vanish, is it necessarily true that $x^{\epsilon}=o(f)$ for every $\epsilon > 0$?
(Note everywhere that I am taking the limit as $x \to \infty$, even though big and little Oh notation applies also to more general limits.)
Motivation: I just realized this simple proof for why the exponential function must grow asymptotically faster than any polynomial $P$.
Assume that $P$ is of degree $n$. Then take the $(n+1)$st term of the exponential's MacLaurin series (denote it $g$); then $P=o(g)$ and $g=O(e^x)$; hence it follows that $P=o(e^x)$.
If $\epsilon$ is non-integer, then just take the ceiling of $\epsilon$ and apply the argument above.
The above question is essentially -- how far can we extend the concepts in this proof? For instance, does the same reasoning allow us to show that $P=o(\Gamma)$ for any $P$? Can we also use a Taylor series proof to show that $e^x=o(\Gamma)$?
I like this proof because it has some nice corollaries.
Natural Logarithm: Since the Taylor expansion of the natural logarithm has no non-vanishing terms, and yet $ln(x) = o(x^\epsilon)$ for every $\epsilon > 0$, it must follow that the radius of convergence for its Taylor series is finite.
Assume this were not true; then a completely analogous proof as for the exponential above would hold, and the logarithm would be super-polynomial. This is a contradiction, since the logarithm is sub-polynomial.
Sine and Cosine: Both of these have MacLaurin series with infinite radii of convergence. Therefore it must follow that for both functions, infinitely many of the terms of their MacLaurin series vanish, or else again we would have a contradiction, since $\sin x=O(1)$ and $\cos x = O(1)$.
This is especially interesting to me because they are based on concepts most commonly applied in computer science and numerics (i.e. discrete mathematics), and yet these are very elegant and appealing proofs for basic calculus (the quintessential discipline of continuous mathematics).
The entire function $f(x)=\sin x + \cos x$ is a counterexample to your statement. All Maclaurin coefficients are nonzero, yet $f$ is bounded on $\mathbf{R}$.