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I can easily prove that $$\sum_{n=1}^{\infty} \frac{z^n}{n^2} $$ uniformly converges when $|z|<1$ simply by applying the M-Test. But, I cannot figure out how to prove that $$\sum_{n=1}^{\infty} \frac{z^n}{n}$$ does not uniformly converge when $|z|<1$. Hope someone could help me out this is simple I know but just can't figure out a way. Thanks

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  • $\begingroup$ @graydad Nope $|z|<1$ $\endgroup$ – Heisenberg Nov 4 '14 at 15:43
  • $\begingroup$ @graydad I just added the other summation to show that I find it easier to prove that a sum is uniformly convergent but find it difficult to prove that it is not when required $\endgroup$ – Heisenberg Nov 4 '14 at 15:44
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    $\begingroup$ Oh I see my mistake; there is a squared term in the first statement. Pardon my beety eyes! $\endgroup$ – graydad Nov 4 '14 at 15:45
  • $\begingroup$ @graydad I those are two DIFFERENT summations $\endgroup$ – Heisenberg Nov 4 '14 at 15:45
  • $\begingroup$ @graydad No worries could happen to anyone $\endgroup$ – Heisenberg Nov 4 '14 at 15:46
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It suffices to show $$\sum_{n=1}^{\infty} \frac{z^n}{n}$$ does not uniformly converge when $0<z<1$

Now consider $\sum_{k=n}^{2n} \frac{z^k}{k}>(n+1)\frac{z^{2n}}{2n}>\frac{z^{2n}}{2}>\frac{1}{4}$, if $(\frac{1}{2})^\frac{1}{2n}<z<1$.

Hence, let $\epsilon=\frac{1}{4}$, for all $n\in\mathbb{N}$,let $(\frac{1}{2})^\frac{1}{2n}<z_0<1$, then $$\sum_{k=n}^{2n} \frac{z_0^k}{k}>(n+1)\frac{z_0^{2n}}{2n}>\frac{z_0^{2n}}{2}>\frac{1}{4}$$

Hence the series is not uniformly convergent on $(0,1)$.

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