Find all complex numbers 
such that the series
![\[ \sum_{n=1}^{\infty} \left( \frac{z}{2z+1} \right)^n \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0b480e23b839c85ee31a69d685786bc4_l3.png "Rendered by QuickLaTeX.com")
converges.
This is a geometric series. It converges if
![\[ \left| \frac{z}{2z+1} \right| < 1 \quad \implies \quad \left| 2 + \frac{1}{|z|} \right| > 1 \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-97727471a95f6ead9502abe652a55347_l3.png "Rendered by QuickLaTeX.com")
and diverges if
![\[ \left| 2 + \frac{1}{|z|}\right| < 1. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-cffa7a552ae511c50275a5c739c5ccf4_l3.png "Rendered by QuickLaTeX.com")
If 
then the series also diverges since 
in this case.
Find all complex numbers such that the series
converges.
This is a geometric series. It converges if
and diverges if
If then the series also diverges since in this case.
What about |2+1/z|=1? Seems like this converges for z≠-1 based on Apostol theorem 10.19 pg. 408.
No, it does not converge. This is actually a very neat point which is important to grasp IMO. The key is that when the absolute value is 1, then the series is BOUNDED, but DOES NOT converge. Using intuition, it would mean that the series x^n, where |x| = 1, x != 1 as a function would resemble something like a sine function: it is bounded by -1/+1, but it never stays at the same value as n grows. Apostol specifically introduced this example to imply this difference, that Dirichlet test can be also used with non-convergent series (which are bounded nevertheless).
Very nice explanation!
Truly, even if the abs. value of the sum is bounded, it can still oscillate inside those bounds.