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

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!

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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 $n\to +\infty$

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thanks Nameless, but i dont get yet, $k_{n}$ as an example for what? for a subsequence of $(-1)^ni^n$? – doniyor Dec 15 '12 at 17:08
@doniyor Yes. Let me add some more detail, – Nameless Dec 15 '12 at 17:09
thanks a lot! nice – doniyor Dec 15 '12 at 17:31

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

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how can i show that divergence ? – doniyor Dec 15 '12 at 16:58
@doniyor: Another answer makes mine complete. Thanks Nameless. – Babak S. Dec 15 '12 at 17:00

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

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