Let $$f(x):=\sum_{n=0}^{\infty} \frac{f^{n}(0)x^n}{n!}$$ where the $$|f^{n}(0)| \le C\frac{\Gamma(\frac{n+1}{\alpha})}{\alpha^{\frac{n+1}{\alpha}+1}}$$
for a constant $C>0$ and $\alpha>0$. Does this imply that the series has a positive radius of convergence? If I assume $\alpha \ge 1$ then the answer is obviously yes, but for $\alpha \in (0,1)$ I am not sure anymore.
For $\alpha=1$ for example, we have $|f^{n}(0)| \le C\ n!$ so that the radius of convergence is $1$.
The main issue here seems to be whether we can bound $\frac{\Gamma(\frac{n+1}{\alpha})}{n!}$ by something that grows not faster than exponentially in $n$ ?
Since $\Gamma(n) \sim cn^{-1/2}(n/e)^n $,
$\begin{array}\\ t_n &=\dfrac{\Gamma((n+1)/a)}{n!a^{(n+1)/a+1}}\\ &\sim \dfrac{c((n+1)/a)^{-1/2}((n+1)/(ae))^{(n+1)/a}}{cn^{1/2}(n/e)^na^{(n+1)/a+1}}\\ &\sim \dfrac{(1/a)^{-1/2}e^{n-((n+1)/a)}((n+1))^{(n+1)/a}}{n(n)^na^{2(n+1)/a+1}}\\ &\sim \dfrac{(1/a)^{-1/2}e^{n(1-1/a)-1/a}((n+1))^{(n+1)/a}}{n^{n+1}a^{2(n+1)/a+1}}\\ &\sim \dfrac{e^{n(1-1/a)}((n+1))^{(n+1)/a}}{a(1/a)^{1/2}e^{1/a}n^{n+1}a^{2(n+1)/a}}\\ &=\dfrac{1}{a^{1/2}e^{1/a}} \dfrac{e^{n(1-1/a)}((n+1))^{(n+1)/a}}{n^{n+1}a^{2(n+1)/a}}\\ &=\dfrac{1}{a^{1/2}e^{1/a}} \dfrac{e^{(n+1)(1-1/a)-1+1/a}((n+1))^{(n+1)/a}}{n^{n+1}a^{2(n+1)/a}}\\ &=\dfrac{1}{a^{1/2}e} \dfrac{e^{(n+1)(1-1/a)}((n+1))^{(n+1)/a}}{n^{n+1}a^{2(n+1)/a}}\\ &=c \dfrac{e^{(n+1)(1-1/a)}((n+1))^{(n+1)/a}}{n^{n+1}a^{2(n+1)/a}}\\ \text{so}\\ t_n^{1/(n+1)} &\sim c^{1/(n+1)} \dfrac{e^{(1-1/a)}((n+1))^{1/a}}{na^{2/a}}\\ &\sim \dfrac{e^{(1-1/a)}((n+1)/n)^{1/a}}{n^{1-1/a}a^{2/a}}\\ &\sim \dfrac{e^{(1-1/a)}}{n^{1-1/a}a^{2/a}}\\ \end{array} $
If $a=1$, this gives $t_n^{1/(n+1)} \sim 1 $, which agrees with OP's statement.
If $a > 1$, then $1-1/a > 0 $, so $t_n^{1/(n+1)} \to 0 $, so the root test says the series converges for all $x$.
If $0 < a < 1$, then $1-1/a \lt 0 $, so that $t_n^{1/(n+1)} \to \infty $, so the root test says the series diverges for all $x$.