An application of Hurwitz's Theorem

Question

Suppose $f(z)=\sum_{n=0}a_nz^n$ such that the radius of convergence is 1 and $a_0\neq0$.Let $s_n(z)=a_0+a_1z+\cdots+a_nz^n$ be the partial sums.Let $1>\delta>0$ such that $f\neq 0$ on $|z|=1-\delta$.Prove that there is $K>0$ and integer N such that for all $n\ge N$ the modulus of the product of zeros of $s_n(z)$ inside $|z|=1-\delta$ is at least $K$.

We know that $f(0)\neq0$,what bothers me is that since for sufficiently large $n$,the zeros of $s_n(z)$ should be near $f$,what if $f$ has no zeros inside $|z|=1-\delta$.The reason to choose a smaller circle than the unit one seems to avoid the zeros of $s_n$ goes to the boundary.I sense that I shall invoke the radius of convergence to derive some constraints on the coefficients so as to ensure polynomial $s_n$ will have zeros inside the circle,which is near zero.

Answer 1

A Hurwitz-based proof may be intuitively clear, but writing it out in every detail takes a bit of work. I'd say: $f$ has distinct zeros $a_1,\dots,a_m$ with multiplicities $k_1,\dots,k_m$. Let $r>0$ be small enough so that the $r$-neighborhoods of the zeros are disjoint and stay inside of $|z|=1-\delta$. Also make sure that $r<\frac12 \min|a_j|$. By Hurwitz, for large enough $n$ the function $s_n$ has $k_j$ zeros inside of the $r$-neighborhood of $a_j$; and it has no other zeroes inside of $|z|=1-\delta$. It follows that the product of its zeroes is at least $$\prod_{j=1}^m (|a_j|-r)^{k_j} \ge \prod_{j=1}^m (|a_j|/2)^{k_j} $$ which is the desired lower bound.

An alternative (in my opinion, better) approach is to use Jensen's formula.

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Remark. The statement could be made cleaner and stronger: for every $\delta\in (0,1)$ there exist $K>0$ and $N$ such that for all $n\ge N$ the modulus of the product of zeros of $s_n$ inside $|z|=1-\delta$ is at least $K$.

Indeed, if $\delta$ happens to be such that $f$ has a zero on the circle $|z|=\delta$, you can use some larger value of $\delta$ instead. There are at most countably many values of $\delta$ to avoid, since $f$ has at most countably many zeros in the unit disk.