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

by RoRi

March 17, 2016

Find all complex numbers $z$ such that the series

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

converges.

Let

$b_n = \frac{1}{2n-1}, \qquad a_n = \left( \frac{1-z}{1+z} \right)^n.$

Then the series converges if $\sum a_n$ converges. But, since $\sum a_n$ is a geometric series, it converges if

$\left| \frac{1-z}{1+z} \right| < 1 \quad \implies \quad |1-z| < |1+z|.$

If we write $z = x + iy$ this inequality holds if and only if $x > 0$ .
On the other hand,

$|1-z| > |1+z| \quad \implies \quad \frac{1-z}{1+z} > 1$

and so

$\lim_{n \to \infty} \frac{1}{2n-1} \left( \frac{1-z}{1+z} \right)^n \neq 0$

hence, the series diverges.

Therefore, the series converges for all $z = x + iy$ with $x \geq 0$ .

Stumbling Robot