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Show that the derivatives of an analytic function cannot satisfy $|f^{(n)}(z)|>n!n^n$ for all $n$ for any $z$ where $f$ is analytic.

My attempt:

Assume otherwise. Say $|f^{(n)}(z)|>n!n^n$.

$a_n$, the $n$th term of the Taylor series for $f$ centered at $z$ is $a_n=\frac{f^{(n)}(z)}{n!}$ and $|a_n|\leq\frac{M}{r^n}$. Thus we have $$n^n<\bigg|\frac{f^{(n)}(z)}{n!}\bigg|=|a_n|\leq\frac{M}{r^n}$$ $$\Rightarrow n^n<\frac{M}{r^n}$$ Which is clearly false for large $n$ but not necessarily for all $n$.

I also tried this:

Assume otherwise. Say $|f^{(n)}(z)|>n!n^n$. Since $f(z)$ is analytic, by Cauchy's theorem $$\bigg|f^{(n)}(z)\bigg|=\bigg|\frac{n!}{2\pi i}\int_{|w-z|=r}\frac{f(w)}{(w-z)^{n+1}}dw\bigg|$$ $$\Rightarrow\frac{1}{2\pi}\bigg|\int_{|w-z|=r}\frac{f(w)}{(w-z)^{n+1}}dw\bigg|>n^n$$ but again I don't know where to go with this.

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    $\begingroup$ Can't you just show that if the assumption is true, hence the Taylor series simply diverges? Which contradicts with the fact that the function is analytic. $\endgroup$ – echzhen Feb 28 '16 at 1:54
  • $\begingroup$ by definition (??), a function is analytic at $a$ if $\sum_{k=0}^\infty \frac{f^{(k)}(a)}{k!} z^k$ has a non-zero radius of convergence $\endgroup$ – reuns Feb 28 '16 at 2:26
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By the post Find the upper bound of the derivative of an analytic function, if $|f(z)|\leqslant M$ on $|z|\leqslant r$, then $$ \bigg|f^{(n)}(z)\bigg|\leqslant \frac{M n!\rho^n}{(r-|z|)^{n}} $$ where $\rho>1$. If $|f^{(n)}(z)|>n!n^n$ for all $n$ for any $z$, let $z=0$ and we have $$ n!\cdot n^n<\frac{M n!\rho^n}{(r-|z|)^{n}}\quad\text{or }\quad (\frac{rn}{\rho})^n<M $$ This is impossible since left is unbounded for large $r$ and $n$.

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