Function that grows faster than all power series

Question

Is there a real valued function that diverges in infinity faster than all power series? I tried looking for some example but all functions I can think of are developable as power series :(

Answer 1

It is an easy construction to show that for any positive (continuos or say just locally bounded) function $\eta(x), x \in \mathbb R$, there is an entire function $F(z)$ with $F(x) >\eta(x), x \in \mathbb R$ (and we can take $F$ without zeroes in the plane so $0 < 1/F < 1/\eta, x \in \mathbb R$ gives an entire function positive on the reals and less than any fixed positive function there too).

Let $M_n =\sup_{|x| \le n+1}\eta(x)+1$ (here we need $\eta$ locally bounded of course, so for example continuos will do).

Pick $k_n \ge n$, s.t $(\frac {n^2}{n+1})^{k_n} \ge M_n, n \ge 1$ and take

$F_1(z) =M_0+\sum_{n \ge 1}(\frac{z^2}{n+1})^{k_n}$

Trivially $F_1$ entire as the series conveges absolutely on any compact set since $k_n \ge n$ and then $F_1(x) >0, x \in \mathbb R$ by construction.

But now $F_1(x) > M_0 >\eta(x), |x| \le 1$ real and if $n \le |x| \le n+1, n \ge 1$ real, we have $(\frac {x^2}{n+1})^{k_n} \ge (\frac {n^2}{n+1})^{k_n} \ge M_n >\eta(x)$ so $F_1(x) > \eta(x)$ there too!

Taking $F(z)=e^{F_1(z)}$ we get $F(x) > F_1(x) > \eta(x), x \in \mathbb R$ and $F$ has no zeroes in the plane!