Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question

3 Answers 3

up vote 4 down vote accepted

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 $n\to +\infty$

share|improve this answer
    
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.

share|improve this answer
    
how can i show that divergence ? –  doniyor Dec 15 '12 at 16:58
2  
@doniyor: Another answer makes mine complete. Thanks Nameless. –  Babak S. Dec 15 '12 at 17:00

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.