# Uniform Convergence of $\sum_{n=1}^{\infty} \frac{z^n}{n}$

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

• @graydad Nope $|z|<1$ – Heisenberg Nov 4 '14 at 15:43
• @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 – Heisenberg Nov 4 '14 at 15:44
• Oh I see my mistake; there is a squared term in the first statement. Pardon my beety eyes! – graydad Nov 4 '14 at 15:45
• @graydad I those are two DIFFERENT summations – Heisenberg Nov 4 '14 at 15:45
• @graydad No worries could happen to anyone – Heisenberg Nov 4 '14 at 15:46

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 $(0,1)$.