Determine the radius of convergence of ∑ zⁿ / 2ⁿ

by RoRi

March 31, 2016

Determine the radius of convergence $r$ of the power series:

$\sum_{n=0}^{\infty} \frac{z^n}{2^n}.$

Test for convergence at the boundary points if $r$ is finite.

Let $a_n = \frac{z^n}{2^n}$ . Then we have

$\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{z^{n+1}}{2^{n+1}} \right) \left( \frac{2^n}{z^n} \right) \\[9pt] &= \frac{z}{2}. \end{align*}$

This implies $r = 2$ . Thus, the series converges for all $z$ such that $|z| < 2$ .
The series converges at none of the boundary points $|z| = 2$ since

$\lim_{n \to \infty} \frac{z^n}{2^n} \neq 0$

if $|z| = 2$ .

Stumbling Robot

Determine the radius of convergence of ∑ zⁿ / 2ⁿ

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