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Does this problem converge?

  1. Jan 30, 2007 #1
    1. The problem statement, all variables and given/known data

    does this problem converge? http://img67.imageshack.us/img67/4246/untitleddg2.jpg [Broken]
    2. Relevant equations

    lim inf.
    n = 1 E cos(pi/n) converge?

    3. The attempt at a solution

    i tried using the squeeze theorem, but thats not the way to do this.

    please help.
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jan 30, 2007 #2


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    Start by thinking about what happens to the individual terms when n gets larger and larger. Do they get bigger or smaller?
  4. Jan 30, 2007 #3
    alright cos(pi/n) approaches 1, as n gets bigger and bigger. I would say it converges to 1?

    no squeeze theorem needed at all?
  5. Jan 31, 2007 #4


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    Go back and read your text! Saying that the sequence converges to 1 certainly does not tell you that the series converges!
  6. Jan 31, 2007 #5
    i don't understand, cos can only oscilate between -1, and 1, so how am i not correct about this. heh.

    it may have been wrong to say it ultimately converges, but i do know it stays between -1, and 1, it may go on for infinity, this could be an example of a harmonic series that diverges, is this what were saying here?

    please help.
  7. Jan 31, 2007 #6
    Actually there's one really simple thing here that I should point out what is [tex]pi/1[/tex] now [tex]pi/2[/tex] now [tex]pi/3[/tex] now [tex]pi/4[/tex] as n increases what happens to the result? and what will eventually happen to the result? What does it therefore converge to? If as n increases it converges to this what can you assume? if as it decreases it does what, what can you assume?

    with something this simple plug some numbers in and see what happens. Works for me:smile:
    Last edited: Jan 31, 2007
  8. Jan 31, 2007 #7


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    Correct about what? Yes, cosine is always between -1 and 1 and [itex]cos(\pi /n)[/itex] converges to 1 as n goes to infinity. The SEQUENCE converges to 1 and that tells you that the SERIES does not converge- in order that a series (sum of a sequence) converge, the sequence must go to 0. It is not an example of a "harmonic series"- that is specifically a series related to 1/n and it has the property that the terms of its sequence, 1, 1/2, 1/3, etc. goes to 0 but the series doesn't converge anyway. Here, the sequence does not converge to 0 (it converges to 1) so you know immediately that the series does not converge.
  9. Jan 31, 2007 #8
    thankyou both for your in depth answers.
  10. Jan 31, 2007 #9

    Where a is a constant.

    General rule: [tex] \lim_{n\rightarrow\infty}[/tex]The cosine of any constant over n will always converge to 1 and therefore converges to cos(0)=1

    [tex]\sum_{n=0}^\infty cos (\frac {a}{n})
    \lim_{n\rightarrow\infty}[/tex] does not converge.

    [tex]\sum_{n=0}^\infty 1 \lim_{n\rightarrow\infty}[/tex]

    1+1...infinity does not converge.

    in fact you could say


    does not converge too since the value 0 is undefined.

    [tex]\sum_{n=0}^\infty cos (\frac {a}{n})
    \lim_{-\infty\rightarrow\infty}[/tex] does not converge.
    Last edited: Jan 31, 2007
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