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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!

3 Answers 3

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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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    @doniyor Yes. Let me add some more detail,2012-12-15
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If $i=\sqrt{-1}$ then the sequence $(i^{3n})_1^{\infty}$ is divergent.

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    @doniyor: Another answer ma$k$es mine complete. Than$k$s Nameless.2012-12-15
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Note that $i^3 = -i$. Hence, $i^{3n} = (-i)^n$. $i^{3n} = (-i)^n = \begin{cases} 1 & n \equiv 0 \pmod{4}\\ -i & n \equiv 1 \pmod{4}\\ -1 & n \equiv 2 \pmod{4}\\ i & n \equiv 3 \pmod{4} \end{cases}$