2
$\begingroup$

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?

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

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.