need help - $\lim_{n \to \infty} i^{3n} $

Question

it seems first easy to me, but now i am tossing my head against wall not being able to solve the problem. i need to check for convergence of this sequence below. i dont know how to start although it seems to be very easy one

$\lim_{n \to \infty} i^{3n} = help = ?$

i need help here. do i have to work with $exp$ here? i seem to have enough material in my brain and cannot use them on time. tragedy!

Answer 1

Write $$a_n=i^{3n}=(i^{3})^{n}=(-i)^{n}=(-1)^ni^n$$ The last limit doesn't exist!

Take $k_n=4n$ and $m_n=4n+2$. Then $(a_{k_n})$ and $(a_{m_n})$ are both subsequences of $(a_n)$ but $$a_{k_n}=(-1)^{4n}i^{4n}=1\cdot 1=1\to 1$$ while $$a_{m_n}=(-1)^{4n+2}i^{4n+2}=1\cdot (-1)=-1\to -1$$ as $n\to +\infty$

Answer 2

If $i=\sqrt{-1}$ then the sequence $(i^{3n})_1^{\infty}$ is divergent.

Answer 3

Note that $i^3 = -i$. Hence, $i^{3n} = (-i)^n$. $$i^{3n} = (-i)^n = \begin{cases} 1 & n \equiv 0 \pmod{4}\\ -i & n \equiv 1 \pmod{4}\\ -1 & n \equiv 2 \pmod{4}\\ i & n \equiv 3 \pmod{4} \end{cases}$$