How bad can a Taylor series be?

Question

My question is, is there a limit to how fast a Taylor series can blow up on the boundary of its radius of convergence? Put formally, given any function $F:(0,1)\rightarrow\mathbb{R}$ such that $\lim\limits_{x\rightarrow 1^-} F(x) = \infty$, does there exist a sequence $a_n\in\mathbb{R}$ such that $g(x) = \displaystyle\sum_{n=0}^\infty a_n x^n$ has radius of convergence 1 and$$ \lim_{x\rightarrow 1^-} \frac{|g(x)|}{F(x)} = \infty? $$ I'm not even sure whether the answer should be yes or no.

Answer 1

After thinking about it further, I came to a solution: The answer is yes. It suffices to show that given an increasing function $F:\mathbb{R} \rightarrow\mathbb{R}$ there exists an entire function $h(x)$ such that $\lim\limits_{x\rightarrow\infty}\frac{h(x)}{F(x)} =\infty$, since an entire function can be transformed into a function with radius of convergence 1 and a singularity at 1 by taking $g(x) = h\left(\frac1{1-x}\right)$. We will assume $F(x)>x$ (the case of $F(x)\le x$ is trivially possible since $h(x)=x$ is entire).

We will use a sparse series $h(z) = \sum\limits_{n=1}^\infty a_n z^{b_n}$ for some increasing sequence of constants $b_n\in\mathbb{N}$ and $a_n >0$. $h$ being entire means that $\lim\limits_{n\rightarrow\infty} a_n^{1/b_n} \rightarrow 0$. For $h$ to outpace $F$, it suffices to have $h(n)\ge nF(n)$ for any $n\in\mathbb{N}$. Thus it suffices to find sequences $a_n, b_n$ such that:\begin{eqnarray} a_n^{1/b_n} &\le& \frac1{\sqrt{n}}\\ a_n n^{b_n} &\ge& nF(n) \end{eqnarray} Since $a_n>0$, it will be convenient to write $a_n = n^{c_n}$ for some real sequence $c_n$. Taking log base $n$ of both inequalities, we are now looking for sequence $c_n,b_n$ such that \begin{eqnarray} \frac{c_n}{b_n} &\le& -\frac12\\ c_n + b_n &\ge& 1+\log_n(F(n)) \end{eqnarray} Rearranging:\begin{eqnarray} c_n +\frac12 b_n &\le& 0\\ c_n + b_n &\ge& 1+\log_n(F(n)) \end{eqnarray} By our assumption that $F(x)>x$, the right hand side of the second inequality is positive. Since the two lines are not parallel, there is a solution for $(b_n,c_n)$ to this system of inequalities with integer $b_n$. Graphically, the solution can be seen to occur in the region of the plane with positive $b_n$. Hence we have there exists an analytic $h$ such that $h(x) > xF(x)$.