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If $|a_n|\leq 1$, for all $n\ge 1$

Then will, $$(1-z)\sum_{n=1}^N a_nz^n$$

Be bounded for all $|z|\leq 1$, with $z\in \mathbb{C}$ and $N\ge 1$

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Is $z$ in $\mathbb C$ or $\mathbb R$? It certainly cant be an integer. – PVAL Jul 28 '13 at 3:43
@PVAL My bad typo, sorry – user46544 Jul 28 '13 at 3:48
I don't think the series will be bounded in general. Take the sequence $a_n=(-1)^{n}$. Then $(1-z)\sum_{n=1}^{N}a_{n}z^{n}$ grows without bound at $z=-1$ in $N$. – user71352 Jul 28 '13 at 3:52

1 Answer 1

$$|(1-z)\sum_{n=1}^N a_nz^n| \leq |(1-z)|\sum_{n=1}^N |z^n|$$ By the triangle inequality, since $|a_n|\leq1$. Now this is clearly bounded if $|z|<1$,because $\sum_{n=1}^\infty |z^n| = \frac1{1-|z|}$ is finite. If $z=1$, then our series is trivially bounded, and if $|z|=1 ,z\ne 1$ then let $a_i=1/z^i$ So our sum becomes $$(1-z)\sum_{n=1}^N 1$$ which is clearly not bounded.

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