Are the coefficients of power series expansion for a real analytic function bounded? Are the coefficients of power series expansion for a real analytic function bounded?
$f(x)=\sum_{n=0}^{\infty}a_n (x-x_0)^n$
We have a sequence $\{a_n\}, n=0,\cdots,\infty$. Is this sequence bounded? Thanks.
 A: If the radius of convergence is greater than $1$, then the sequence must be bounded, because we would have
$$
\sum_{k=0}^{\infty}a_{k} = f(x_{0}+1) < \infty
$$
since $x_{0}+1$ lies within the interval of convergence. Since the series of coefficients converges, the sequence of coefficients itself must converge to $0$, which implies that the sequence is bounded.
On the other hand, if the radius of convergence is less than or equal to $1$, the coefficients may be unbounded. Just consider 
\begin{align*}
f(x) = \sum_{k=0}^{\infty}(k+1)x^{k} &= \frac{d}{dx}\left[\sum_{k=0}^{\infty}x^{k}\right] \\
&= \frac{d}{dx}\left[\frac{1}{1-x}\right] \\
&= \frac{1}{(1-x)^{2}}.
\end{align*}
A: Consider $f(x) = \frac{1}{1 - 2x}$ expanded at $x = 0$. Then $f(x) = \sum 2^n x^n$, so $a_n = 2^n$ is unbounded.
If the radius of convergence is greater than 1, or at least converges at (at least one of) $x_0 \pm 1$, then the terms are guaranteed to be bounded. This is simply because the terms in a convergent sum must go to zero, hence $|a_n| \to 0$, from which you can conclude that $a_n$ is bounded.
A: The sequence 
$\frac{\log|a_n|}{n}$ is bounded, i.e. there exist $C$, $\rho>0$ so that
$|a_n| \le C \cdot \rho^n$ for all $n\ge 0$. 
Eg: $(2^n \cdot n^3 )$ yes, $(n!)$ no.
