Show that $f(n) = \left(\frac{\sqrt 2}{2}(1+i)\right)^n$ diverges I am having this sequence:
$$f (n)= \left( \frac{\sqrt 2}{2}(1+i)\right)^n$$
I think it is divergent, because I found a subsequence that is divergent:
The subsequence $4n$ shows that the sequence alternates between $1$ and  $-1$.

Questions:


*

*Is this correct?

*How can I write my reflections down in a formally correct way?


$$\frac{\sqrt2}2(1+i)=\frac1{\sqrt2}+\frac1{\sqrt2}i=e^{\frac\pi4i}\implies$$
$$f(n)=e^{\frac{n\pi}4i}=\begin{cases}1&,\;\;n=0\pmod 8\\\frac1{\sqrt2}(1+i)&,\;\;n=1\pmod8\\i&,\;\;n=2\pmod8\\\frac1{\sqrt2}(-1+i)&,\;\;n=3\pmod 8\\-1&,\;\;n=4\pmod 8\\-\frac1{\sqrt2}(1+i)&,\;\;n=5\pmod 8\\-i&,\;\;n=6\pmod 8\\\frac1{\sqrt2}(1-i)&,\;\;n=7\pmod 8\end{cases}$$
Thus, as $\;n\to\infty\;,\;\;f(n)\;\;$  approaches nothing: it keeps on attaining all the eight values above.
