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Homework Help: Does this series converge or diverge?

  1. Apr 11, 2012 #1
    1. The problem statement, all variables and given/known data


    3. The attempt at a solution

    I really couldn't figure this one out. It's definitely not telescoping series because I don't see repeating terms that cancel. I would try to compare it to a geometric series. so what I did was factoring out a 4^n:


    replacing (2+sinn-(1/2+sinn)) with "a"

    Then I noticed the bounds 1<2+sinn<3

    so the whole thing inside the bracket is oscillating between 0<a<8/3

    So i concluded that the series is divergent? (highly unsure) because the coefficient that 4^n multiplies by (by coefficient I mean "a") is always either 0 or a positive number. So when it is 0, there will be no effect. When it is a positive number, the sums will add up, and the series will eventually go to infinity.. Eh.. is that.. a valid line of reasoning?

    Thanks in advance,

  2. jcsd
  3. Apr 11, 2012 #2
    Since 4n→∞ as n→∞, the (2+sin n) - 1/(2+sin n) part had at least better go to 0. Can you show whether the limit of that part is 0 as n→∞?
  4. Apr 11, 2012 #3
    I am quite sure the limit of the bracket part doesn't go to zero because sin(n) is always oscillating.

    But is it supposed to?
    Last edited: Apr 11, 2012
  5. Apr 11, 2012 #4
    L'Hopital's rule doesn't apply since we don't have either of the indeterminate forms 0/0 or ∞/∞ (usually we can't use it when we have trig functions in the fraction like that), so we'll have to try something else.

    2+sin n is definitely oscillating, but we can't really assume either way that the -1/(2+sin n) would make it tend to 0 or oscillate still.
    Still, if you look at a graph, it looks just like a sine or cosine graph that's been translated and stretched, but I have no idea how you could write it as just one trig function...
  6. Apr 11, 2012 #5
    You can show that the derivative of (2+sin x) - 1/(2+sin x) oscillates, so (2+sin n) - 1/(2+sin n) doesn't tend to 0, and so the series diverges.
  7. Apr 11, 2012 #6
    Got it, thank you very much
  8. Apr 11, 2012 #7


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    The derivative of a function does not have to go to zero for the function itself to go to zero

    Also, It's not enough to show that the continuous version of the function oscillates. For example [itex] \lim_{x\to \infty} cos(2 \pi x)[/itex] does not exist, but [itex] \lim _{n \to \infty} cos(2\pi n) = 1[/itex].

    The argument to show that 2+sin(n)-1/(2+sin(n)) does not go to zero is only slightly non-trivial. Suppose that this is close to zero. Then 2+sin(n) is approximately 1/(2+sin(n)) so it must be that 2+sin(n) is approximately 1 and sin(n) is approximately -1. Hence it must be that n is very close to 3pi/2+2kpi for some k. Based on this prove that 2+sin(n+1)-1/(2+sin(n+1)) cannot be close to zero
  9. Apr 11, 2012 #8
    I see. Thanks for the reply.

    One additional question: Can you solve this question by using any of the following techniques? These are the techniques we learned in class, so it is most probable that our professor wants to test us on one of them, but it could be an anomaly.

    - Comparison with geometric
    - Comparison with p-series
    - Ratio test
    - Root test
    - Divergence test
    - Alternating series
    - Limit comparison
  10. Apr 11, 2012 #9
    O_S, I see that we can't just assume as you showed with that example.
    The derivative in this case turned out to be (1+1/(sinx+2)2)cosx. The part in front is bounded above by 2 and bounded below by 10/9, so what else could be done to show that this doesn't settle on any number?
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