Radius of Convergence of $\sum_{n=0}^{\infty}n^2z^n$

Question

Find the radius of convergence of $$\sum_{n=0}^{\infty}n^2z^n$$

(If it matters, $z$ is a complex variable.)

My attempt:

The radius of convergence is

\begin{align*} R&=\lim\limits_{n\to\infty}\Bigl\lvert\dfrac{a_n}{a_{n+1}}\Bigl\lvert\\ &=\lim\limits_{n\to\infty}\Bigl\lvert\dfrac{n^2z^n}{(n+1)^2z^{n+1}}\Bigl\lvert\\ &= \dfrac{1}{z} \end{align*}

I was under the impression that the radius of convergence was supposed to be a number, so am I doing something wrong, or is this correct?

Answer 1

Apply the general result:

If $\lim \frac{a_{n+1}}{a_n} = r$ exists, and $r<1$ then $\sum |a_n|$ converges, and if $r>1$ $\sum |a_n|$ diverges.

Here, $$ \frac{a_{n+1}}{a_n} \to z $$and then

• if $|z|<1$ there is convergence, and then $R\ge 1$
• if $|z|>1$ there is divergence, and then $R\le 1$

hence the radius $R= 1$