How to evaluate $ \lim_{n\to \infty} \frac{i}{n}\left(\frac{1+i}{\sqrt{2}}\right)^n $ where $i=\sqrt{-1}$?

Question

How to evaluate the following limit? $$ \lim_{n\to \infty} \frac{i}{n}\left(\frac{1+i}{\sqrt{2}}\right)^n $$ Here $i=\sqrt{-1}$.

I got: $$\lim_{n\to \infty} \frac{i}{n}\left(\frac{1+i}{\sqrt{2}}\right)^n = \lim_{n\to \infty} \frac{(i-1)^n}{n(\sqrt{2})^n} $$ I know the lower part goes to infinity but what to do with the upper part? Is that usefull to use squeeze theorem or is there any simplier way?

Answer 1

We have $$ \left|\frac{1+i}{\sqrt{2}}\right|=1 $$ Thus $$ \left|\frac{i}{n}\left(\frac{1+i}{\sqrt{2}}\right)^{\!n}\right|=\frac{1}{n} $$

Answer 2

$$\lim_{n\to \infty} |\frac{i}{n}\left(\frac{1+i}{\sqrt{2}}\right)^n|=\lim_{n\to \infty} {1\over n}=0$$therefore $$\lim_{n\to \infty} \frac{i}{n}\left(\frac{1+i}{\sqrt{2}}\right)^n=0$$