Let $f$ be the a power series over $\mathbb{C}$ of infinite radius. The order $\rho(f)$ of $f$ is defined to be the infinum of the following set : $$ \{ A \ge 0 \mid \exists r_0\ge 0, \forall r\ge r_0, M_f(r) \le \exp(r^A)\}$$ Where $M_f(r)$ is the supremum of $f$ over all complex numbers of norm $r$.
How does one show the following identify : $$ 1/\rho(f) = \lim\inf (- \dfrac{\ln|a_n|}{n\ln n})$$ Where $f(z) = \sum_{n\ge 0} a_n z^n$
I managed to show an inequality. Let $A > \rho(f)$, then for $r$ big enough, $M_f(r) \le \exp(r^A)$. We then use the identity $$2\pi r^n a_n = \int_0^{2\pi} f(re^{it}) e^{-int}\mathrm{d}t $$ To obtain $$ |a_n| \le r^{-n}M(r)$$ For $r$ big enough, we therefore have $$ | a_n|\le \exp( r^A - n\ln r)$$ $$ \rightarrow \dfrac{-\ln |a_n|}{n\ln n} \ge \frac{\ln(r)}{\ln n} - \dfrac{r^A}{n\ln n}$$ For $n$ big enough, one can set $r= n^{1/A}$ and use the previous inequality : $$ \dfrac{-\ln |a_n|}{n\ln n} \ge \frac{1}{A} - \frac{1}{\ln n}$$ Taking the lower limit, we obtain $$\lim\inf (- \dfrac{\ln|a_n|}{n\ln n}) \ge \frac{1}{A}$$ Then, by letting $A\to \rho(f)$ I get $$\lim\inf (- \dfrac{\ln|a_n|}{n\ln n}) \ge \dfrac{1}{\rho(f)} $$
I'm having trouble showing the other side now, my idea was to take $A < \lim\inf (- \dfrac{\ln|a_n|}{n\ln n})$, then there exists $n_0\ge 0$ such that $n\ge n_0$ implies $$ -\dfrac{\ln |a_n|}{n\ln n} \ge A $$ $$ \rightarrow \ln |a_n| \le - A n\ln n$$ $$ |a_n| r^n \le \exp( n\ln r - A n\ln n)$$ $$ |a_n| r^n \le \left( \frac{r^{1/A}}{n}\right)^{nA} $$ At first sight, this looks good as the right hand side is close to $$ \dfrac{ r^{n/A}}{n^n} \le \dfrac{r^{n/A}}{n!} $$ And the latter sums to $\exp(r^{1/A})$ and the other inequality follows. But I don't see how I can achieve that inequality. I tried taking $B< A$ but it doesn't get rid of the exponent over the $n^n$. I also tried splitting the sum in 2 according to $ r^{1/A} /n \le \ge 1$ and use inequality but the splitting depends on $n$. So I am stuck, and starting to feel like I should try to pursue another idea, even though this seems to be the "natural" path. But I can't think of anything.
Any ideas to solve this problem ? Thanks.
You're on a good way, just the final estimate is missing. This is easier to do with the whole series than termwise. Let me however replace your $A$ with $1/\vartheta$ so that at the end I get exponents of $\vartheta$ instead of $1/A$. What we want to show is $$\frac{1}{\vartheta} < \liminf_{n \to \infty} \biggl(\frac{-\ln \lvert a_n\rvert}{n\ln n}\biggr) \implies \vartheta \geqslant \rho(f)\,.$$ You already found $$\lvert a_n\rvert < \frac{1}{n^{n/\vartheta}}$$ for $n \geqslant n_0$. Hence there is a $b_1 > 0$ such that $$\lvert a_n\rvert \leqslant b_1\cdot \frac{1}{n^{n/\vartheta}}$$ holds for all $n$. Then we have $$M(r) \leqslant \sum_{n = 0}^{\infty} \lvert a_n\rvert r^n \leqslant b_1\cdot \sum_{n = 0}^{\infty} \frac{r^n}{n^{n/\vartheta}}$$ for all $r \geqslant 0$. For fixed $r > 0$ the expression $$\frac{r^t}{t^{t/\vartheta}}$$ attains its maximum at $t = r^{\vartheta}/e$, and the maximum is $$\exp \biggl(\frac{r^{\vartheta}}{e}\log r - \frac{r^{\vartheta}}{e} \log \frac{r}{e^{1/\vartheta}}\biggr) = \exp \biggl(\frac{r^{\vartheta}}{e\vartheta}\biggr)\,.$$ We use this bound for the terms with index $n < (2r)^{\vartheta}$. For $n \geqslant (2r)^{\vartheta}$ we have $$\biggl(\frac{r}{n^{^/\vartheta}}\biggr)^n \leqslant \frac{1}{2^n}$$ and thus we obtain \begin{align} M(r) &\leqslant b_1\cdot\sum_{n = 0}^{\infty} \frac{r^n}{n^{n/\vartheta}} \\ &\leqslant b_1\biggl(\bigl(1 + (2r)^{\vartheta}\bigr)\exp \biggl(\frac{r^{\vartheta}}{e\vartheta}\biggr) + 2\biggr) \\ &\leqslant 4b_1(2r)^{\vartheta}\exp \biggl(\frac{r^{\vartheta}}{e\vartheta}\biggr) \end{align} for $r \geqslant 1/2$. This implies $\vartheta \geqslant \rho(f)$, as needed.