Find the radius of convergene R for power series

Question

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?

Answer 1

$\sum e^n z^n = \sum (ez)^n$ 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 $e>1$.

Answer 2

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 $z=\frac{-1}{e}$