# $\sum_{n=1}^\infty\frac{z^n}{n}$ does not converge uniformly on $\mathbb{D}$.

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?

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

## 1 Answer

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

• 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? – Idonotknow Feb 5 at 17:12
• @Idonotknow And there are $n$ terms in the sum. – Julián Aguirre Feb 5 at 17:39
• so which is more stronger? – Idonotknow Feb 5 at 18:07
• $n\times\frac1n=1$. – Julián Aguirre Feb 5 at 18:09
• so it will never be greater than ? so why we are using greater than or = and not = only? – Idonotknow Feb 5 at 18:29