# 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.

• Thanks for drawing our attention to samuelj.li/complex-function-plotter
– lhf
Commented Nov 26, 2021 at 14:05
• The limit expression is only valid for $|z|<1$. Also note that in the complex numbers, there is no order, so "$-1<z<1$" makes only sense for real numbers. Commented Nov 26, 2021 at 14:40
• You can simplify this example by removing a few superfluous terms: $z^n-1$ converges to $-1$. Where do the roots go? Commented Nov 26, 2021 at 14:40
• @Arthur it converges to -1 iff $|z|<1$. For $|z|=1$ we have $1^\infty$, but can we manage to solve the limit as $n\to\infty$ to get that the infinite roots lie on the unit disk? Commented Nov 26, 2021 at 14:47
• @Peter you're right. Edited in $|z|<1$ Commented Nov 26, 2021 at 14:50

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.

• This is great! My numerical testing suggests that apart from that single zero $\approx 1/2$ the rest of them are actually outside of the unit circle (unless you leave the $z-1$ uncancelled which then produces a single zero on the unit circle). Any thoughts about how to see that (if true)? Commented Nov 26, 2021 at 20:54
• Those zeros satisfy $$z^n+\frac1z=2.$$ For real $x$ the function $x^n+1/x$ is increasing when $x>n^{-1/(n+1)}$. If $\epsilon$ rules out zeros below that threshold, it follows that $|z|\ge1$. Commented Nov 26, 2021 at 21:22
• @JyrkiLahtonen: If you try Rouché on the circle $|z|=1$ instead of $|z|=1-\epsilon$, you still get $|h|<|g|$ on the whole circle, except at the point $z=1$, and I think (although I haven't thought very carefully about it) that it might be possible to get $|h|<|g|$ on the contour consisting of the unit circle except with an arbitrary small indentation going just inside $z=1$. If that's correct, the $n$ “large” zeros would indeed be one at $z=1$ and the rest outside the unit circle. Commented Nov 26, 2021 at 22:00