I've got a problem to solve. It relates to my homework and I couldn't manage to deal with it.
Prove that $$ \sum_{n=1}^{\infty} \frac{1}{n^2} z^{-2n} $$ is uniformly continuous for $\lbrace z:|z|>1 \rbrace$.
I suspect, that i should use theorem from lecture:
- (Bolzano-wieierstrass theorem): $D \in C$, if terms of the series $$\sum_{n=0}^{\infty} f_n(z) , \sum_{n=0}^{\infty} a_n$$ satisfy condition: $$\forall_{n\in \mathbb{N}}\text{ }\forall_{z\in \mathbb{C}}\text{ }|f_n(z)|\leq a_n$$ Then $\displaystyle\sum_{n=0}^{\infty} f_n(z)$ is uniformly continuous, when $\displaystyle\sum_{n=0}^{\infty} a_n$ is continous.
Maybe I'am wrong and more "traditional" way is better:
$$ \forall_{\epsilon>0}\text{ } \exists_{N(\epsilon)}\text{ } \forall_{n\in \mathbb{N}}\text{ } \forall_{z\in \mathbb{D}}\text{ } [n>N(\epsilon) \Longrightarrow |S(z)-S_{n}(z)|<\epsilon]$$
Where: $D\in \mathbb{C}$, $S(z)$- sum of series, $S(z)= \displaystyle\lim_{n \rightarrow \infty} S_{n}(z)$.
By M-Test the series converges uniformly for $\{|z| \geq 1\}$. So $f(z)= \sum_{n=1}^{\infty} \frac{1}{n^2} z^{-2n} $ is a continuous function there. Besides$|f(z)| < \sum_{n=1}^{\infty} \frac{1}{n^2} R^{-2n} <\sum_{n=1}^{\infty} R^{-2n}=\frac {R^{-2}} {1-R^{-2}}$ for $|z| >R$ and $\frac {R^{-2}} {1-R^{-2}} \to 0$ as $R \to \infty$. Hence, $f(z) \to 0$ as $|z| \to \infty$. This implies that $f$ is uniformly continuous on $\{|z| \geq 1\}$.