Show that
\[ \sum_{n=0}^{\infty}\frac{z^n}{(n!)^\alpha} \]
is an entire function of order \(1/\alpha\).
Solution
Let \[ F(z)=\sum_{n=0}^\infty \frac{z^n}{(n!)^\alpha},\qquad \alpha>0. \] The ratio test shows that the radius of convergence is infinite, so \(F\) is entire.
Use the elementary Stirling lower bound \(n!\ge (n/e)^n\). Put \(R=r^{1/\alpha}\). Then \[ \frac{r^n}{(n!)^\alpha} \le \left(\frac{eR}{n}\right)^{\alpha n}. \tag{1} \] For real \(x>0\), the function \[ x\bigl(1+\log R-\log x\bigr) \] has derivative \(\log(R/x)\) and therefore has maximum \(R\) at \(x=R\). Hence every term on the right of (1) is at most \(e^{\alpha R}\).
Let \(N=\lceil 2eR\rceil\). For \(n>N\), \(eR/n<1/2\), so (1) gives \[ \frac{r^n}{(n!)^\alpha}\le 2^{-\alpha n}. \] Consequently \[ M_F(r)\le F(r) \le (N+1)e^{\alpha R}+\sum_{n>N}2^{-\alpha n} \le C e^{C'R} =C e^{C'r^{1/\alpha}} \] for suitable constants \(C,C'>0\). Hence the order is at most \(1/\alpha\).
For the reverse inequality, suppose \(F\) had order \(\rho<1/\alpha\). Choose \(\sigma\) with \(\rho<\sigma<1/\alpha\). Then for some \(A,B>0\), \[ M_F(r)\le Ae^{Br^\sigma}. \] Cauchy’s estimate gives \[ \frac1{(n!)^\alpha}\le \frac{Ae^{Br^\sigma}}{r^n}. \] Taking \(r=(n/(B\sigma))^{1/\sigma}\) minimizes the right-hand side, so \[ \log (n!)^\alpha \ge \frac n\sigma\log n+O(n). \] But Stirling gives \(\log (n!)^\alpha=\alpha n\log n+O(n)\), forcing \(\alpha\ge1/\sigma\), contrary to \(\sigma<1/\alpha\). Thus the order is exactly \[ \boxed{1/\alpha}. \]