Determine for which values of $z$ the series $\sum_{n=1}^{\infty} 2^{n}n^{n}z^{n}$ converges For which values of 
$z$ does $\sum_{n=1}^{\infty} 2^{n}n^{n}z^{n} $ converge?
I know the first step is to perform a ratio test to find the radius of convergence, but I'm having trouble choosing an $a_{k}$ to find the ratio between   $a_{k}$ and  $a_{k+1}$ .
 A: If ratio test is the first step, have a look at the zeroeth step: Does the summand tend to $0$ at all?
If $z\ne0$ then for $n$ large enough $|nz|>1$ and hence $|2^nn^nz^n|>2^n$.
A: You can use the ratio test with $a_n=2^n n^n$.
However to get a better understanding of power series, I would suggest to go to the basics. If $\sum_{n=1}^{\infty} 2^{n}n^{n}z^{n} $ converges, then the term of the series has to be bounded. Which means that you can find $M >0$ such that $$\vert 2^{n}n^{n}z^{n} \vert < M$$ for all $n \in \mathbb N$, or $\vert z \vert < \frac{M^{1/n}}{2n}$. As the RHS tends to $0$, the radius of convergence is equal to $0$.
A: The root test gives $\sqrt[n]{|2^{n}n^{n}z^{n}|}=|2nz| \to \infty$ and so the series diverges, unless $z=0$.
A: Look at the term 
For $z$ such that $|z|<1,$ $$|2^n n^nz^n|=|(2n)^nz^n|\ge(2n)^n\frac 1 {n^n}=2^n>1\text{ for }n>\frac 1{|z|}$$
For $z$ such that $|z|\geq 1$, $$|2^n n^nz^n|\ge(2n)^n>1$$
And for $z=0$, the series is just zero.
Hence, the series is converges only at $z=0$.
