# Prove that partial sums of $\sum_{n=1}^{\infty}{z^n}, z \in \mathbb{C}, |z|=1$ are bounded

It's obvious that if $arg(z)=2\pi \cdot q,\ q \in \mathbb{Q},\ q=\frac{a}{b}$ then $\sum_{n=1}^b{z^n}=0$. Thus the partial sums are bounded by $\sum_{n=1}^{b-1}{z^n}$.

But what to do with the case of irrational argument?

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Don't forget the case $z = 1$ :) Else, it's only the formula for a geometric progression ... – Selim Ghazouani Mar 21 '12 at 19:09
Also, you don't know that the sum is bounded by $\sum_{n=1}^{b-1} z^n$, only that it is bounded by the maximum of $\sum_{n=1}^k z^n$ for $k=1,2,...b-1$ – Thomas Andrews Mar 21 '12 at 19:26
"But what to do with the case of irrational argument?" Use $\sum_{n=0}^{N-1} q^n = \frac{1-q^N}{1-q}$ for $q\ne 1$. – Sam Mar 21 '12 at 19:28
@Selim, thank you, it's my stupid. – Sergey Filkin Mar 22 '12 at 5:38
@Thomas, you're right. – Sergey Filkin Mar 22 '12 at 5:45

Assume $z\neq 1$ (if $z=1$, then the partial sums are not bounded.)

Then $s_k = \sum_{n=1}^k z^n = z\frac{z^{k}-1}{z-1}$.

If $z$ is not a root of unity (that is, if $z=e^{2\pi i \alpha}$ for some irrational $\alpha$) then the supremum of $|s_k|$ is $\frac{2|z|}{|z-1|}$. Basically, $z^k$ can be, at furthest, a distance of $2$ away from $1$, and, when $\alpha$ is irrational, we can mak $z^k$ arbitrarily close to $-1$.

When $\alpha=\frac{a}{b}$ (in reduced form,) the value is:

$$\sup |s_k| = \frac{|z(e^{2\pi ix}-1)|}{|z-1|}$$

where $x = \frac{\lfloor \frac{b}{2}\rfloor}{b}$.

When $b$ is even, then $x=\frac{1}2$ and we get the same max as we got for irrational $\alpha$.

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The partial sums are of the form $a_n+i\cdot b_n$ where $$a_n=\sum_{k=1}^n \cos k \theta, \ a_n=\sum_{k=1}^n \cos k \theta$$ where $\theta \in (0,2\pi)$. In this wikipedia article, you can find the exact expressions for $a_n,b_n$ and see that they are bounded.

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