Behavior of the roots of an infinite series.

I have the polynomial $P_n(z)=1-\sum_{k=1}^{n}z^k$. We know that this polynomial has exactly $n$ roots in $\mathbb{C}$. Let $\rho$ be the number of roots of $P_n$, thence if $n\to\infty$ then $\rho$ must tend to $\infty$ too. Though, if we interpret the sum as a geometric series, we get that $$P_n(z)=1-\sum_{k=1}^{n}z^k=\frac{z^{n+1}-2z+1}{1-z}$$ And if we make $n\to\infty$ it only converges for $|z|<1$, becoming $P_\infty(z)=-\frac{2z-1}{1-z}$, that has only one real root for $z=\frac{1}{2}$. So, where did the other roots go? I plotted $P_n$ variating $n$ and noted that the roots tend to accumulate on the unit disk. (Interactive Mapping). So, can we say that all the roots will accumulate on the unit disk as $n\to\infty$ for $z\neq1$? How can we prove this? Thanks!

Here I have some screenshots of the mapping. Respectively, $n=5$, $n=10$, $n=100$. You can see that the roots tend to accumulate towards the unit circle.

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Thanks.

Solution 1:

This can be proved using Rouché's theorem.

Let $0 < \epsilon < 1/2$, and take $N$ large enough that $$ (1-\epsilon)^{N+1} < 1-2\epsilon \quad\text{and}\quad (1+\epsilon)^{N+1} > 3 + 2 \epsilon . $$ Then for any $n \ge N$ we have $$ (1-\epsilon)^{n+1} < 1-2\epsilon \quad\text{and}\quad (1+\epsilon)^{n+1} > 3 + 2 \epsilon $$ too, and I will show that this implies that the annulus $1-\epsilon \le |z| < 1+\epsilon$ contains all roots of $f(z) = 2z-1-z^{n+1}$ except $z=1/2$.

Let $$ g(z) = 2z-1 = 2(z-\tfrac12) ,\qquad h(z) = -z^{n+1} . $$ Then on the circle $|z|=1-\epsilon$, the point closest to $1/2$ is $z=1-\epsilon$, so we have $$ |g(z)| = 2 |z-\tfrac12| \ge 2 \bigl((1-\epsilon) - \tfrac12 \bigr) = 1 - 2 \epsilon , $$ and $|h(z)|=(1-\epsilon)^{n+1}$, so according to the inequalities above we have $|h(z)|<|g(z)|$ on the circle, which according to Rouché means that $f=g+h$ has equally many zeros inside the circle as $g$, namely one (the only zero of $g$ is at $1/2$).

On the other hand, on the circle $|z|=1+\epsilon$, the point furthest from $1/2$ is $z=-1-\epsilon$, so we have $$ |g(z)| = 2 |z-\tfrac12| \le 2 \bigl|(-1-\epsilon) - \tfrac12 \bigr| = 3 + 2 \epsilon , $$ and $|h(z)|=(1+\epsilon)^{n+1}$, so according to the inequalities above we have $|h(z)|>|g(z)|$ on the circle, which according to Rouché means that $f=g+h$ has equally many zeros inside the circle as $h$, namely $n+1$ (since $h$ has a zero of that multiplicity at the origin).

Thus, for any $n \ge N$, the function $f$ has $n$ of its $n+1$ zeros in the annulus between those two circles, of radius $1 \pm \epsilon$. And $\epsilon$ can be chosen arbitrarily small to begin with, which shows that the zeros do accumulate on the unit circle.