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Does the complex sequence z_n=n/(1+i)^n converge?

  1. May 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Determine whether the sequence zn = n/((1+i)n) converges and rigorously justify your answer.

    2. Relevant equations


    3. The attempt at a solution
    I have attempted an ε-n proof using my limit as 0 (as exponentials grow faster than polynomials I assumed this was the correct limit), and I said |(1+i)n| = |1+i|n = √2 n but I was stuck with how to show n/(√2 n) < ε.

    The only other way I could think to do this was to show the sequence was increasing (or decreasing) and bounded above (or below) but I wasn't sure how to do that either.

    Can someone please point me in the right direct? I find complex sequences quite confusing as a concept.
     
  2. jcsd
  3. May 3, 2015 #2

    Mark44

    Staff: Mentor

    This seems difficult to me as well, due to n showing up as an exponent and in the numerator.
    The ratio test might be helpful here. ##\lim_{n \to \infty} \frac{|z_{n + 1}|}{|z_n|}## works out to a value less than 1, which implies that the magnitudes of your sequence values are decreasing.
     
  4. May 3, 2015 #3

    pasmith

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    Homework Helper

    Taking logs of [itex]n/2^{n/2} < \epsilon[/itex] yields [tex]
    \log n - \frac{n}{2} \log 2 < \log \epsilon.[/tex] Now if [itex]\epsilon[/itex] is small and positive then [itex]\log \epsilon[/itex] is large and negative, so the question is: if [itex]R > 0[/itex] is arbitrary, can you always find an [itex]n \in \mathbb{N}[/itex] such that [tex]
    \frac{\log 2}{2}n - \log n > R?[/tex]
     
  5. May 3, 2015 #4

    Ray Vickson

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    Do you mean literally what you wrote, viz., that you want to know if ##z_n = n/(1+i)^n## converges/diverges, or did you really mean to ask about convergence/divergence of the series ##\sum_n z_n##? In either case, try writing ##1+i ## in polar form.
     
  6. May 3, 2015 #5
    Thanks guys,
    Mark44 - I don't think I can use the ratio test because it is a test for if series converge.
    pasmith - I have never seen that method before :/

    What if i split the sequence into its real and imaginary parts then show that each of those converge which will imply the sequence converges?
     
  7. May 3, 2015 #6
    Ray Vickson - I meant literally does the sequence z_n converge
     
  8. May 3, 2015 #7

    Mark44

    Staff: Mentor

    The ratio I wrote shows that the sequence |zn| is decreasing, which was one of the things you mentioned in your first post.
     
  9. May 3, 2015 #8

    Mark44

    Staff: Mentor

    I think that he (Tommy941) is working with polar form, Ray, which is where he's getting ##(\sqrt{2})^n##. That's my take at any rate.
     
  10. May 3, 2015 #9
    Ahh Mark44 I see what you mean, so if I can show z_n is decreasing using the ratio test, then I can say it converges to 0 (as the sequence is clearly bounded below by 0).
    I was also thinking could I do this by writing the denominator in polar form, rationalising it, then splitting the sequence into real and imaginary parts and show that each part converges to some limit (im guessing 0) and then the limit of z will be the sum of these two limits?
     
  11. May 3, 2015 #10

    Mark44

    Staff: Mentor

    I don't think this is a good idea. In polar form it will be of the form ##re^{\text{something}i}##. The exponential part gives a complex number somewhere on the unit circle. It's not too hard to show using what I suggested that the magnitudes (|zn|) are a decreasing sequence. What you're doing is trapping these complex numbers in smaller and smaller circles.
     
  12. May 3, 2015 #11

    Ray Vickson

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    Right: I missed that.
     
  13. May 3, 2015 #12

    WWGD

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    Gold Member

    You can use that ## \frac {n}{ \sqrt 2^n}= \frac {n}{2^{n/2}} ## Then n doesn't stand a chance against ##2^{n}## (which is a shift of ## 2^{n/2}## . You can use a L'Hopital-style argument for more rigor.
     
    Last edited by a moderator: May 3, 2015
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