Consider a power series $\sum_{n=0}^{\infty} a_n x^n$ that is convergent for all real x, thus defining a function $f: \mathbb{R} \to \mathbb{R}$. Can one give necessary and sufficient criteria the sequence of the coefficients $(a_n)$ has to meet in order for $f$ to be bounded on $\mathbb{R}$? (Let's disregard the trivial case that $a_0$ is the only non-zero coefficient and let's call a sequence "function-bounded" if the power series is bounded.) Criteria for boundedness seem to be far more difficult to obtain than the usual criteria for convergence of a power series, here some remarks:

a) A necessary condition for $\sum_n a_n x^n$ to be bounded is that there are infinitely many non-zero coefficients which change sign infinitely many times.

b) The boundedness of $f$ is an "unstable" property of the sequence of coefficients: any non-zero change in any finite subset (except $a_0$) will destroy boundedness. Thus the linear subspace of all function-bounded sequences is rather "sparse" in the vector space of all sequences representing convergent power series.

c) On the other hand, the linear subspace of all function-bounded sequences contains at least all power series of functions that can be written as $\cos \circ h$ with $h$ an entire, real-analytic function, and the algebraic span of these functions. One could conjecture that this space is already the space of all bounded functions that can be represented as power series[EDIT: seems to be refuted, cf. comment below]. And perhaps this could be a starting point for deducing the criteria.

EDIT (conjecture added):
Is is true, that every power series $\sum_{n=0}^{\infty} a_n x^n$ that is convergent for all real $x$ can be modified only by changing the signs of the terms to a convergent power series $\sum_{n=0}^{\infty} \epsilon_n a_n x^n, \quad \epsilon_n \in \{\pm1\}$ that is bounded for all real $x$?

Example: One can modifify the signs of the power series of the exponential function $\sum_{n=0}^{\infty} x^n/n!$ pretty easily to a bounded power series by $\epsilon_n = +1$ for $n = 0 or 1 \mod 4$ and $\epsilon_n = -1$ for $n = 2 or 3 \mod 4$, yielding the function $\sin(x) + \cos(x)$. (One can modify the signs pretty easily a bit further such that the power series is not only bounded on the real axis, but also on the imaginary axis - but this is not the question here).
I have neither succeeded in finding a counterexample nor in prooving this conjecture.

EDIT2: Thanks for the nice counterexample. I would like to improve the conjecture as follows: Define a power series $\sum_{n=0}^{\infty} a_n x^n$ as nondominant, if for all $x \in \mathbb{R}$ the absolute value of every term $a_kx^k$ is smaller or equal than the sum of the absolute values of all the other terms: $|a_kx^k| \le \sum_{n \neq k} |a_n x^n|$. The improved conjecture is: Is is true, that every nondominant power series $\sum_{n=0}^{\infty} a_n x^n$ that is convergent for all real $x$ can be modified only by changing the signs of the terms to a convergent power series $\sum_{n=0}^{\infty} \epsilon_n a_n x^n, \quad \epsilon_n \in \{\pm1\}$ that is bounded for all real $x$?

nondominant seriestherein implies it is zero (consider by contradiction the first non-zero term $a_k$ and let $x\to0$). A weaker definition of nondominance may lead to interesting results though. $\endgroup$ – Pietro Majer Mar 10 '12 at 8:45