Show that $f(n) = \left(\frac{\sqrt 2}{2}(1+i)\right)^n$ diverges

Question

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 $-1$.

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Questions:

1. Is this correct?
2. How can I write my reflections down in a formally correct way?

Answer 1

$$\frac{\sqrt2}2(1+i)=\frac1{\sqrt2}+\frac1{\sqrt2}i=e^{\frac\pi4i}\implies$$

$$f(n)=e^{\frac{n\pi}4i}=\begin{cases}1&,\;\;n=0\pmod 8\\\frac1{\sqrt2}(1+i)&,\;\;n=1\pmod8\\i&,\;\;n=2\pmod8\\\frac1{\sqrt2}(-1+i)&,\;\;n=3\pmod 8\\-1&,\;\;n=4\pmod 8\\-\frac1{\sqrt2}(1+i)&,\;\;n=5\pmod 8\\-i&,\;\;n=6\pmod 8\\\frac1{\sqrt2}(1-i)&,\;\;n=7\pmod 8\end{cases}$$

Thus, as $\;n\to\infty\;,\;\;f(n)\;\;$ approaches nothing: it keeps on attaining all the eight values above.

Answer 2

It is correct. Since $$|f(4n+4)-f(4n)|=2\quad\forall n\in\Bbb N$$ the sequence is not a Cauchy sequence, and hence it diverges.

Answer 3

The number $\sqrt{2}(1+i)/2$ has complex modulus 1 and angle $\pi/4$. Thus, the sequence $f(n)$ dances around the origin along the vertices of a regular octagon.

Answer 4

Note that $$f(n) = \left( \frac{\sqrt[]{2}}{2}(1+i)\right)^n =\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\cdot i\right)^n = \left(e^{i\pi/4}\right)^n = e^{i\pi n/4}$$

Hence $$f(4n) = e^{i\pi n} = (-1)^n$$ which diverges when $n\to\infty$.