Home » Blog » Find all z such that the series (z / (2z+1))n converges

Find all z such that the series (z / (2z+1))n converges

Find all complex numbers z such that the series

    \[ \sum_{n=1}^{\infty} \left( \frac{z}{2z+1} \right)^n \]

converges.


This is a geometric series. It converges if

    \[ \left| \frac{z}{2z+1} \right| < 1 \quad \implies \quad \left| 2 + \frac{1}{|z|} \right| > 1 \]

and diverges if

    \[ \left| 2 + \frac{1}{|z|}\right| < 1. \]

If \left| 2 + \frac{1}{|z|} \right| = 1 then the series also diverges since \lim_{n \to \infty} a_n \neq 0 in this case.

2 comments

    • Artem says:

      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).

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):