Find all complex numbers 
such that the series
![\[ \sum_{n=1}^{\infty} n^n z^n \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-110870d103af9132e95ed88e49c40cfb_l3.png "Rendered by QuickLaTeX.com")
converges.
We consider the limit of the 
th root of the terms of the series,
![\begin{align*} \lim_{n \to \infty} (a_n)^{\frac{1}{n}} &= \lim_{n \to \infty} \left( n^n z^n \right)^{\frac{1}{n}} \\[9pt] &= \lim_{n \to \infty} nz \\[9pt] &= \begin{dcases} +\infty &\text{if } z \neq 0 \\ 0 & \text{if } z = 0. \end{dcases} \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-a5b31e42388c4c3a5df05741e771320a_l3.png "Rendered by QuickLaTeX.com")
Hence, the series converges if and only if 
.
Find all complex numbers such that the series
converges.
We consider the limit of the th root of the terms of the series,
Hence, the series converges if and only if .
Is it the root test used here? I do not think that is legal, since the root test is defined only for real numbers in Apostol (it requires a sequence of non-negative terms, which nz clearly does not follow).
Hello!
Yes, I think it’s a mistake too, as an alternative it can be proved that except when
the necessary condition of convergence 
is not satisfied.
as 
goes to 
.
When z=0, if we want to be pedantic, we can show that the limit of partial sums is
Cheers!
But if you consider absolute convergence, it is positive and it is real. Because absolute convergence implies the conditional one, the proof seems valid to me.