Domain of convergence of $\sum n e^{-n}z^n$ .

Question

Let $\sum n e^{-n}z^n$ be a power series. To find the radius of convergence and domain of convergence of the power series....

I have found the radius of convergence to be $e$ by ratio test. Also seen that at $z=e$ the series diverges.

Can we conclude from here that the domain of convergence is the open ball $|z| < e$?

Answer 1

From what you've done so far... no, you haven't got enough to conclude that the domain of convergence is the open ball of radius $e$. The boundary behavior might be more subtle; consider, for example, a similar series like

$$\sum_{n = 1}^{\infty} \frac{z^n}{n}$$

whose domain of convergence includes $z = -1$, although the radius is $1$.

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However, in this particular case, you're ok. Note that if $|z| = e$ then

$$|n e^{-n} z^n| = n \left(\frac{|z|}{e}\right)^n = n \to \infty$$

and so the series diverges on the entire boundary.