Evaluate $ \lim_{n\to\infty} \frac{n^n}{3^n} a_n$ For
$$ f(z)= \sum_{n=0}^{\infty } a_n z^n  $$
with
$$ |f(z)| \leq M e^{|z|},  $$
calculate
$$ \lim_{n\to\infty} \frac{n^n}{3^n} a_n$$
Tried several methods but I don't know how to approach this correctly, thanks for your time.

This is for $f$ being an entire function. By Cauchy's formula
$$
a_n  = \frac{1}{{2\pi i}}\oint_{\left| t \right| = r} {\frac{{f(t)}}{{t^{n + 1} }}dt} ,
$$
with some $r>0$. Thus,
$$
\left| {a_n } \right| \le \frac{1}{{2\pi }}\oint_{\left| t \right| = r} {\frac{{\left| {f(t)} \right|}}{{\left| t \right|^{n + 1} }}\left| {dt} \right|}  \le \frac{1}{{2\pi }}\oint_{\left| t \right| = r} {\frac{{Me^{\left| t \right|} }}{{\left| t \right|^{n + 1} }}\left| {dt} \right|}  = \frac{1}{{2\pi }}\oint_{\left| t \right| = r} {\frac{{Me^r }}{{r^{n + 1} }}\left| {dt} \right|}  = M\frac{{e^r }}{{r^n }}.
$$
Since $f$ is entire, $r$ can take any positive value. For $a_n$, take $r=n$. Hence,
$$
\frac{{n^n }}{{3^n }}\left| {a_n } \right| \le \frac{{n^n }}{{3^n }}M\frac{{e^n }}{{n^n }} = M\left( {\frac{e}{3}} \right)^n  \to 0,
$$
since $e/3<1$.
