Zn = ((1+i)/squareRoot(3))^n Zn = n(1/i)^n Please Help...Not sure how to do this...
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Originally Posted by jzellt Zn = ((1+i)/squareRoot(3))^n Zn = n(1/i)^n Please Help...Not sure how to do this... What are you trying to do? I assume from the title, that you are taking limits, but the limit as n approaches what? And are we finding limits of the sequences or are we summing them?
Sorry... Find the limit as n approaches infinity of each sequence that converges. If the sequence diverges, explain why.
Originally Posted by jzellt Zn = ((1+i)/squareRoot(3))^n Zn = n(1/i)^n Please Help...Not sure how to do this... Can you go from here?
Last edited by Prove It; February 22nd 2010 at 11:48 PM.
Originally Posted by jzellt Zn = ((1+i)/squareRoot(3))^n Zn = n(1/i)^n Please Help...Not sure how to do this... So . Can you go from here?
Originally Posted by Prove It Can you go from here? CB
Originally Posted by CaptainBlack CB Yes it is, thanks. Will edit.
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