Two equal power series with different centres Suppose $a_n$ for $n\geq 0$ is a complex number such that $$\sum_{n\geq 0}a_n(z-2i)^n=\sum_{n\geq 0}(n+1)z^n$$ holds for  each $z\in \mathbb{C}$ such that $|z|<1$ and Im$(z)>0$ (imaginary part). Then show that $$\lim\sup\sqrt[n]{|a_n|}=\frac1{\sqrt{5}}.$$
I know that for the series $\sum_{n\geq 0}(n+1)z^n$, radius of convergence is $1$. This means the same is also true for the series given in left hand side. But how to connect this with $\lim\sup\sqrt[n]{|a_n|}$?
 A: A two-step approach. At first we transform the series $\sum_{n\geq 0}(n+1)z^n$ to determine $a_n$. Then we calculate the $\limsup$.

We obtain
\begin{align*}
\sum_{n\geq 0}(n+1)z^n
&=\frac{d}{dz}\left(\sum_{n\geq 0} z^{n+1}\right)=\frac{d}{dz}\left(\frac{z}{1-z}\right)\\
&=\frac{1}{(1-z)^2}\\
&=\frac{1}{\left(\left(1-2i\right)-\left(z-2i\right)\right)^2}\\
&=\frac{1}{(1-2i)^2}\sum_{n\geq 0}\binom{-2}{n}\left(-\frac{z-2i}{1-2i}\right)^n\tag{1}\\
&=\frac{1}{(1-2i)^2}\sum_{n\geq 0}\color{blue}{\frac{n+1}{(1-2i)^n}}(z-2i)^n\tag{2}\\
\end{align*}

Comment:

*

*In (1) we make a binomial series expansion.


*In (2) we use the binomial identity $\binom{-2}{n}(-1)^n=\binom{n+1}{n}=n+1$.

From (2) we derive
\begin{align*}
\color{blue}{\limsup_{n\to \infty}\sqrt[n]{|a_n|}}
&=\limsup_{n\to \infty}\sqrt[n]{\left|\frac{n+1}{(1-2i)^n}\right|}\\
&=\limsup_{n\to \infty}\sqrt[n]{\frac{n+1}{|1-2i|^n}}\\
&=\frac{1}{|1-2i|}\limsup_{n\to \infty}\sqrt[n]{n+1}\\
&=\frac{1}{|1-2i|}\\
&\,\,\color{blue}{=\frac{1}{\sqrt{5}}}
\end{align*}
and the claim follows.

