Zn = ((1+i)/squareRoot(3))^n
Zn = n(1/i)^n
Please Help...Not sure how to do this...
$\displaystyle \lim_{n \to \infty}\left(\frac{1 + i}{\sqrt{3}}\right)^n = \lim_{n \to \infty}\left[\frac{\sqrt{2}}{\sqrt{3}}\left(\cos{\frac{\pi}{4}} + i\sin{\frac{\pi}{4}}\right)\right]^n$
$\displaystyle = \lim_{n \to \infty}\left[\left(\frac{\sqrt{6}}{3}\right)^n\left(\cos{\frac{ n\pi}{4}} + i\sin{\frac{n\pi}{4}}\right)\right]$
$\displaystyle = \lim_{n \to \infty}\frac{(\sqrt{6})^n}{3^n} \cdot \lim_{n \to \infty}\left(\cos{\frac{n\pi}{4}} + i\sin{\frac{n\pi}{4}}\right)$
Can you go from here?
$\displaystyle z_n = n\left(\frac{1}{i}\right)^n$
$\displaystyle = n\left(\frac{i}{i^2}\right)^n$
$\displaystyle = n\left(-i\right)^n$
$\displaystyle = n\left[\cos{\left(-\frac{\pi}{2}\right)} + i\sin{\left(-\frac{\pi}{2}\right)}\right]^n$
$\displaystyle = n\left[\cos{\left(-\frac{n\pi}{2}\right)} + i\sin{\left(-\frac{n\pi}{2}\right)}\right]$
So $\displaystyle \lim_{n \to \infty}z_n = \lim_{n \to \infty}n\left[\cos{\left(-\frac{n\pi}{2}\right)} + i\sin{\left(-\frac{n\pi}{2}\right)}\right]$
$\displaystyle = \lim_{n \to \infty}n \cdot \lim_{n \to \infty}\left[\cos{\left(-\frac{n\pi}{2}\right)} + i\sin{\left(-\frac{n\pi}{2}\right)}\right]$.
Can you go from here?