# Show that $\sum \frac{z^n}{n}$ diverges if $z = 1$ but otherwise converges if $|z|=1$.

Hi: I'm reading John D'Angelo's textbook "An Introduction To Complex Analysis and Geometry" and trying ( emphasis on trying ) to work on the exercises in Chapter 4. I'm already stuck on only the second one. The question is:

Show that $\sum \frac{z^{n}}{n}$ diverges if $z = 1$ but otherwise converges if $|z| = 1$.

I think I might have trouble with each of the exercises and there are 13 more. So, if anyone knows of a existing solution manual for the text, please let me know. I'm not a student so just trying to learn this on my own. Otherwise, I'll just keep trying one per day and posting to this list when I'm stuck. Thanks a lot.

• Do you mean "otherwise converges if $|z| < 1$"? Or am I missing something? – Darth Geek Aug 15 '14 at 18:07
• @DarthGeek The series given converges for all $|z| = 1$ bar when $z=1$. As for why (to the OP), the most to a hint I can give is to use summation by parts: en.wikipedia.org/wiki/Summation_by_parts – Andrew D Aug 15 '14 at 18:08
• @AndrewD Wow, this result is really interesting! I had no idea! – Darth Geek Aug 15 '14 at 18:09
• The Dirichlet test for convergence is rather useful (for example here). I suggest taking a look at that. – Daniel Fischer Aug 15 '14 at 18:13
• The series given seems to be the Taylor series for $-\ln(1-z)$. That would explain the convergence for $|z| = 1$ except for $z = 1$. – Darth Geek Aug 15 '14 at 18:13

You can use Abel's complex test (a generalization of Abel's test). It says that if $$\sum\limits_{n=0}^{+\infty}a_n z^{n}$$ converges when $|z|<1$ and diverges when $|z|>1$, then when $a_n$ decreases monotonically to zero, the series converges on $|z|=1$ everywhere except $z=1$
Hint: The Dirichlet test was mentioned in the comments. Note that $1 + z + z^2 + \ldots + z^N = (z^{N+1} - 1)/(z-1)$, and $z \neq 1$ has magnitude 1.
• Hi: I looked up the Dircihlet test and the requirements for $a_{n}$ are satisfied. But I don't know what M should be so that the third requirement: $|\sum_{i=1}^{N} b_{i}| < M$ is satisfied. Thanks. – mark leeds Aug 15 '14 at 22:32
• Hi: The looked up the Dirchlet test but I don't what M can be so that $|\sum_{1}^{N}| < M$. – mark leeds Aug 15 '14 at 22:33