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$f_n(z)=\frac{z^n}{n}$, where it is uniformly convergent?

well, $\frac{z^n}{n}\to 0\forall |z|\le 1$ am I right?

Myshkin
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It certainly is uniformly convergent in $D(0,1)$ as $|\frac{z^n}{n}-0| \leq \frac{1}{n}$, so $\forall \epsilon > 0$ and $\forall z \in D(0,1)$ select n such that $\frac{1}{n} < \epsilon$. Which we can guarentee exists by Archimedian property. Hence, the result follows. Select any number slightly greater than 1 in modulus.... what can you conclude?

ADA
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