Complex power series exercise

Question

I'm stuck in a the following point of an exercise. Consider the following series $$\sum_{n = 1}^\infty \left(1 + \frac{i}{n}\right)^{n^3} z^n,$$

it says that the convergence radius is $\frac{1}{\sqrt{e}}$. How can I prove that?

Answer 1

Note that we have

$$\begin{align} \lim_{n\to\infty}\left| a_n\right|^{1/n}&=|z|\,\lim_{n\to\infty}\left|\left(1+\frac in\right)^{n^2} \right|\\\\ &=|z|\,\lim_{n\to\infty}\left(1+\frac1{n^2}\right)^{n^2/2}\\\\ &=|z|e^{1/2} \end{align}$$

Answer 2

We can use the Hadamard criterion. So:

$$\frac{1}{\rho}=\lim_{n\rightarrow \infty} \sqrt[n]{a_n},$$

in your case we have: $$\frac{1}{\rho}=\lim_{n\rightarrow \infty} \sqrt[n]{\Big|\Big(1+\frac{i}{n}\Big)^{n^3}\Big|}=\lim_{n\rightarrow \infty}\Big| \Big(1+\frac{i}{n}\Big)^{n^2}\Big|=\lim_{n\rightarrow \infty}\Big| \Big(1+\frac{i}{n}\Big)\Big|^{n^2}=\lim_{n\rightarrow \infty}\Big| \Big(\frac{n+i}{n}\Big)\Big|^{n^2}=\lim_{n\rightarrow \infty}\Big(\sqrt{\frac{n^2+1}{n^2}}\Big)^{n^2}=\lim_{n\rightarrow \infty}\Big( \Big(1+\frac{1}{n^2}\Big)^{n^2}\Big)^{\frac{1}{2}}=\sqrt{e}.$$

So $\rho=\frac{1}{\sqrt{e}}$.