# Find the radius of convergene R for power series

For power series, find the radius of convergence R and determine if it is conditionally convergent, absolutely convergent, or divergent for $$z = R$$ and $$z = −R$$.

$$\sum_{i=0}^{\infty} e^n z^n$$

I'm trying to do root test, I think it is divergent as $$C > 1$$ but how do I find the radius of convergence R?

• Do you know the Hadamard Formula? – Lucas Corrêa Sep 29 '18 at 2:14
• certainly $|ez|<1$ has some solutions... – David Peterson Sep 29 '18 at 2:17

## 2 Answers

$$\sum e^n z^n = \sum (ez)^n$$ is a geometric series and so converges iff $$|ez|<1$$. Therefore, $$R=1/e$$.

For $$z=\pm R$$, we get $$\sum e^n(\pm 1)^n/e = \frac1e \sum (\pm e)^{n}$$, which diverges absolutely because $$e>1$$.

Consider $$\sum_{n=0}^{\infty} a_nz^n$$

$$\lim_{n \to \infty} \biggr \rvert \frac{a_{n+1}}{a_n} \biggr \rvert = L$$ Then radius of convergence $$R=1/L$$

What is $$L$$?

$$a_{n+1}=e^{n+1}$$ and $$a_n=e^n$$ so $$L=e$$ and hence radius of convergence is $$\frac{1}{e}$$

Can you go on from here? Check what happens when $$z=\frac{1}{e}$$ and when $$z=\frac{-1}{e}$$