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A serie covergent or not?

  1. Feb 18, 2009 #1
    This is not an homework, but just a couriosity that I have
    I never found a easy way to solve the question of convergence of this serie

    [tex]\sum_n (\sin(n))^n[/tex]

    any help is appreciated!
  2. jcsd
  3. Feb 18, 2009 #2


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    look at exp(in)=cos(n)+isin(n)
    Now because exp(in^2) converges also sin(n)^n converges.

    I hope I haven't done a mess with the question here.
  4. Feb 18, 2009 #3


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    Based on a maple plot I have made, it appears not to converge.
    Plotted is
    [tex]f(X) = \sum_{n=1}^{n=X}\sin{n}^n[/tex]
    from X=0,100000

    Attached Files:

  5. Feb 24, 2009 #4
    Or, i guess we could have also reasoned this way:

    SInce |sin(n)|<1 for all n positive integers. then the series

    [tex]\sum_{n=1}^{\infty}|sin(n)|^n [/tex] converges, so the original series converges absolutely=> it also simply converges.
  6. Feb 24, 2009 #5
    I wish we could! :)

    think of
    [tex]\sum_{n=1}^{\infty} \left( 1 - {1 \over n} \right)^n[/tex]
  7. Feb 24, 2009 #6
    The proof I wrote actually shows it DOES NOT converge.

    I can't proof it diverges to +infinity yet.

    My proof is very tecnical and LONG and actually there are some points a bit difficoult that I have to clear out before to call an EXACT proof.

    I need help!
  8. Feb 24, 2009 #7

    I can't get why exp(in^2) converges.

    Can you explain, please?
  9. Feb 24, 2009 #8
    I wish too!:tongue:

    I wrote before i gave it a thought!:redface:
  10. Mar 1, 2009 #9
    Please help me find this simpler limit, no summation. Probably I knew before but not now.

    \left( 1 + {a \over n} \right)^n
    as n goes to infinity, and a is a constant.
  11. Mar 1, 2009 #10
    That is a famous limit, you can find it's value by taking a logarithm and using L'Hopital's rule.
  12. Mar 1, 2009 #11


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    log(1+a/n)/n=a{log(1+a/n)/(a/n)} -> a{1}=a

    log(1+x)/x -> log'(1)=1
  13. Mar 2, 2009 #12
    [itex]\sum_{n=1}^\infty (sin(n)^n[/itex] diverges.

    (please excuse my english, i'm not familiar with math in english, but I hope you'll understand me anyway)

    with a sum : [tex]S = \sum^{\infty} u_n[/tex], if S is convergeant, then [tex]\lim_{n \to \infty} u_n = 0[/tex].
    by contraposition we can say that if [tex]\lim_{n \to \infty} u_n \ne 0[/tex] then S is not convergeant, and so it is divergeant.

    In this case :
    and [tex]-1\le sin(n)\le1[/tex], comment : we can't use a strict inegality here
    so [tex]-1\le sin(n)^n\le1[/tex],
    [tex]\lim_{n \to \infty} u_n \ne 0[/tex],
    and so S is divergeant.
    Last edited by a moderator: Mar 3, 2009
  14. Mar 2, 2009 #13
    Actually, you know, I agree with wouzzat. I think he's right... any other thoughts?
  15. Mar 2, 2009 #14
    Isn't zero (0) also somewhere between -1 and 1? I have forgotten, pardone my bad memory!:tongue:
  16. Mar 3, 2009 #15
    I believe that what he's saying is that no matter how far you go out, it will always keep hitting '1' again every 2PI steps... '-1' every once and a while, it never really approaches zero in any meaningful way.
  17. Mar 3, 2009 #16
    Well, that might be the case, it is just that i don't think we can jumb to that conclusion solely from what is in there, since like i said, 0 is still a possibility to be eventually the limit of that function; wierd stuff happens as n-->infty. we need some other way to show that. But you are right, we can see that it does not approach 0 at any finite nr n, but like i said, that is just a sign to prove that it doesn't approach 0, it is not a proof on itself.
  18. Mar 3, 2009 #17


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    You don't actually go out 2Pi steps in sin(n). In fact, it's fairly easy to see that sin(n) will NEVER be 1 or -1 for n a natural number. The argument doesn't conclude what the limit of sin(n)n as n goes to infinity will be

    loop's argument is wrong because ein^2) has absolute value one, so summing over it doesn't converge. In fact, if you can tie in the sin series with the cosine equivalent (cos(n)^n) so that one converges if and only if the other does (not sure if you can though) then this would be a proof that it doesn't converge.

    I'd give my solution, but it's super top secret and doesn't exist :p
  19. Mar 3, 2009 #18
    Good point, Office Shredder. You're right, of course...
  20. Mar 3, 2009 #19


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    Probably a dead end but... the question of whether sin(n)n actually goes to 0 can changed into a question of how fast you can approximate pi by rationals (since sin(n) will be close to 1 or -1 when n is close to a multiple of pi, say k*pi, so n/k is close to pi). I distinctly remember something in my algebraic number theory course about algebraic/transcendental numbers being classified based on how quickly you could converge to them based on how large the denominator is, so I'll look that up and see if it helps with anything.

    Regardless, I don't think this is the intent of how the problem was supposed to be solved
  21. Mar 3, 2009 #20
    Hum... yes, actually I didn't proved anything except that "maybe the limit is 0, maybe it isn't".

    I'm curious about the final word of that...
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