TITLE: $f$ entire but not polynomial then $\lim_{n\to\infty}\sup \{|z|:p_n(z)=0\}\to\infty$, where $p_n$ is $n-th$ Taylor series of $f$
QUESTION [5 upvotes]: Let $f(z)=\sum_{k=0}^{\infty}a_kz^k$ be entire and not a polynomial. Let $p_n(z)=\sum_{k=0}^n a_kz^k$ be its $n-th$ Taylor polynomial centered at $0$, and let $r_n=\sup \{|z|:p_n(z)=0\}$. Show that $\lim_{n\to\infty}r_n=\infty$ 
My thought:
$\lim_{n\to\infty}p_n(z)=\lim_{n\to\infty}(z-a_1)(z-a_2)...(z-a_n)=f(z)$, then if $\lim_{n\to\infty}r_n\neq 0$, then this means the zeros of $f$ is bounded by a compact set, then ${z: f(z)=0}$ has a limit point in $\mathbb{C}$, so $f\equiv 0$, which is contradiction with $f$ is polynomial.
I feel something is wrong in my proof above, but I can not tell what exactly is not right.
Could someone kindly help me with this? Thank you so much!

REPLY [8 votes]: Suppose all the roots of $p_n(z)$ are $z_1,z_2,\ldots,z_n$. Vieta's formula implies
$$|z_1 z_2 \ldots z_n|=\left|\frac{a_0}{a_n}\right|$$
So $r_n\geq\sqrt[n]{|z_1 z_2\ldots z_n|}=\sqrt[n]{\left|\frac{a_0}{a_n}\right|}$. Since $f$ is entire, $\lim\sup\sqrt[n]{|a_n|}=0$. The conclusion follows.