Prove that a sequence has no limit

Question

Given $Z_n=\arg{\frac{i^n}{n}}$, how do I show that it has no limit?

Answer 1

Recall that

• $i^2=-1$
• $i^3=-i$
• $i^4=1$
• $i^5=i$

then we have that

• for $n=4+4k$ even $Z_n=\arg{\frac{i^n}{n}}\to 0$

• for $n=2+4k$ even $Z_n=\arg{\frac{i^n}{n}}\to \pi$

Answer 2

HINT: Observe that $Z_n$ takes only four values.