Q2 TUT 0301

[Victor Ivrii]

Determine whether the given infinite series converges or
diverges:
\begin{equation*}
\sum_{n=1}^\infty \Bigl(\frac{1+2i}{\sqrt{6}}\Bigr)^n.
\end{equation*}

[Meng Wu]

By Ratio Test:
$$\begin{align}\sum_{n=1}^\infty \Biggl(\frac{1+2i}{\sqrt{6}}\Biggl)^n &= \lim_{n\to\infty}\Biggl|\frac{\Big(\frac{1+2i}{\sqrt{6}}\Big)^{n+1}}{\Big(\frac{1+2i}{\sqrt{6}}\Big)^{n}}\Biggr|\\&=\lim_{n\to\infty}\Biggl|\frac{1+2i}{\sqrt{6}}\Biggr|\\&=\Biggl|\frac{1+2i}{\sqrt{6}}\Biggr|\\&=\sqrt{\Big(\frac{1}{\sqrt{6}}\Big)^2+\Big(\frac{2}{\sqrt{6}}\Big)^2}\\&=\sqrt{\frac{5}{6}}<1\end{align}$$
Therefore,  the given infinite series converges.