Is this the correct radius of convergence of the following series

Question

Find the radius of convergence of the following series

$$\sum_{n=0}^\infty ni^nz^n$$

Using $$\frac1R = \limsup_{n \to\infty} |a_n|^{1/n}, \quad \text{where }a_n=ni^n,$$ I found $R=1$.

Answer 1

Correct, another way is to differentiate the geometric series $$\frac1{1-w} = \sum_{n=0}^\infty w^n$$ getting $$ \frac{1}{(1-w)^2} = \sum_{n=1}^\infty nw^{n-1} = \frac{1}{w} \sum_{n=1}^\infty nw^n, $$ which is equivalent to $$\frac{w}{(1-w)^2} = \sum_{n=1}^\infty nw^n.$$

Now let $w = iz$ to obtain the closed form, and the manipulation is valid in the radius of convergence of the original geometric series, so for $1 > |w| = |iz|$, which holds iff $|z|<1$.

Answer 2

A ratio test does it: $$ \left|\frac{a_{n+1}}{a_n}\right| = \left| \frac{(n+1)i^{n+1}z^{n+1}}{ni^nz^n} \right| = |z|\frac{n+1} n \to |z| \text{ as } n\to\infty. $$ So the series converges if $|z|<1$ and diverges if $|z|>1.$ (One need not examine what happens when $|z|=1$ in order to know the radius of convergence.)