The limit of complex sequence $$\lim\limits_{n \rightarrow \infty} \left(\frac{i}{1+i}\right)^n$$
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

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