It is required to prove that the series $\sum_{n=1}^\infty\frac{z^n}{n}$ does not converge uniformly on the open unit disc centered at $0$, i.e. $\mathbb{D}$. Clearly by virtue of the ratio test the series converges pointwise on $\mathbb{D}$. However I find it difficult to see why the convergence is not uniform. Could someone please give me a hint?

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    $\begingroup$ What is the set $\mathbb D$? $\endgroup$ – hamam_Abdallah Dec 11 '16 at 15:20
  • $\begingroup$ The open unit disc centered at zero. $\endgroup$ – Janitha357 Dec 11 '16 at 15:22
  • $\begingroup$ Then you might want to include it in your post. $\endgroup$ – Jack Dec 11 '16 at 15:23
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    $\begingroup$ If the summation index is $i$, shouldn't there be an $i$ somewhere in the sum? $\endgroup$ – user940 Dec 11 '16 at 15:38
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    $\begingroup$ The series will diverge at $z=1$ so one expects to have problems as we approach this value. $\endgroup$ – Piotr Benedysiuk Dec 11 '16 at 15:48

Use the uniform Cauchy test. Given $n\in\mathbb{N}$ $$ \Bigl|\sum_{k=n+1}^{2n}\frac1n\,\Bigl(1-\frac1n\Bigr)^k\Bigr|\ge\Bigl(1-\frac1n\Bigr)^{2n}\to e^{-2}>0. $$

  • $\begingroup$ how do you know that the sum on the right greater than the quantity on the left? does not there is 1/n term that lessen each term of the sum? $\endgroup$ – Idonotknow Feb 5 at 17:12
  • $\begingroup$ @Idonotknow And there are $n$ terms in the sum. $\endgroup$ – Julián Aguirre Feb 5 at 17:39
  • $\begingroup$ so which is more stronger? $\endgroup$ – Idonotknow Feb 5 at 18:07
  • $\begingroup$ $n\times\frac1n=1$. $\endgroup$ – Julián Aguirre Feb 5 at 18:09
  • $\begingroup$ so it will never be greater than ? so why we are using greater than or = and not = only? $\endgroup$ – Idonotknow Feb 5 at 18:29

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