The limit of complex sequence

Question

$$\lim\limits_{n \rightarrow \infty} \left(\frac{i}{1+i}\right)^n$$

I think the limit is $0$; is it true that $\forall a,b\in \Bbb C$, if $|a|<|b|$ then $\lim\limits_{n\rightarrow \infty}\left(\frac{a}{b}\right)^n=0$?

I would like to see a proof, if possible. Thank you

Answer 1

You are right. For a proof, just observe that if $\frac{|a|}{|b|}<1$, then $$ \left\|\left(\frac ab\right)^n\right\| = \left(\frac{|a|}{|b|}\right)^n \to 0 $$ as $n\to\infty$.

Answer 2

$\frac{i}{1+i} = \frac{i(1-i)}{(1+i)(1-i)} = \frac{i(1-i)}{2}=\dfrac{e^{\frac{\pi}{2}i}e^{-\frac{1}{4}\pi i}}{\sqrt{2}} = \dfrac{e^{\frac{\pi}{4}i}}{\sqrt{2}}$, so $(\frac{i}{1+i})^n = \dfrac{e^{\frac{n\pi}{4}i}}{(\sqrt{2})^n}$

Answer 3

Observe that $$\frac{i}{1+i} =\frac{1}{2}+\frac{i}{2}$$ Now define $$F_n=\Big(\frac{i}{1+i}\Big)^n$$ Then $$F_1=\frac{1}{2}+\frac{i}{2}$$ $$F_2=\frac{i}{2}$$ $$F_3=-\frac{1}{4}+\frac{i}{4}$$ $$F_4=-\frac{1}{4}$$ $$F_5=-\frac{1}{8}-\frac{i}{8}$$