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Thread: Challenge problem ( Infinite series)

  1. #1
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    Challenge problem ( Infinite series)



    Can you do it ?
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  2. #2
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    i have some opinions about the second one

    sum from k=2 to infinity a^(logb k)

    It is equal to sum from k=2 to infinity k^(logb a)
    = sum from k=2 to infinity k^-(logb 1/a)
    = sum from k=1 to infinity k^-(logb 1/a) - 1
    = Z[ logb(1/a) ] - 1 where Z is Riemann Zeta Function
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  3. #3
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    Quote Originally Posted by simplependulum View Post
    i have some opinions about the second one

    sum from k=2 to infinity a^(logb k)

    It is equal to sum from k=2 to infinity k^(logb a)
    = sum from k=2 to infinity k^-(logb 1/a)
    = sum from k=1 to infinity k^-(logb 1/a) - 1
    = Z[ logb(1/a) ] - 1 where Z is Riemann Zeta Function

    That's a really good step.
    from here
    = sum from k=2 to infinity k^-(logb 1/a)

    A even better step is to recoqnize this as a P-series which converges for

    logb a <-1
    lnb < -lna
    b< 1/a
    ab<1
    .

    Good job though now someone should try the second one.

    It's a tough one.
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by ╔(σ_σ)╝ View Post

    Can you do it ?
    The answer to your question is "no, I can't".

    $\displaystyle \sum_{k=2}^{\infty}\left(\frac{\sin k +2}{3}\right)^k$

    All I can say is that the root test shows that it converges, because $\displaystyle \frac{1}{3}<\lim_{k\to\infty}\frac{\sin k +2}{3}<1$

    Beyond that, I can't say much.
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    Quote Originally Posted by redsoxfan325 View Post
    The answer to your question is "no, I can't".

    $\displaystyle \sum_{k=2}^{\infty}\left(\frac{\sin k +2}{3}\right)^k$

    All I can say is that the root test shows that it converges, because $\displaystyle \frac{1}{3}<\lim_{k\to\infty}\frac{\sin k +2}{3}<1$

    Beyond that, I can't say much.
    It's not an easy question elementary test learned in ordinary textbooks fail to break this question.

    The as k ---> infinity sink --> we don't know .

    You arrived at
    $\displaystyle \frac{1}{3}<\lim_{k\to\infty}\frac{\sin k +2}{3}<1$

    But what if sin K = 1

    Then the root test fails. Because the limit =1


    I think this question reduces to finding out if there is a finite number of times such that sink = 1 for every k E I
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  6. #6
    Super Member redsoxfan325's Avatar
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    $\displaystyle \sin k\neq 1~\forall k$ because $\displaystyle k\in\mathbb{N}$ and thus is never an odd multiple of $\displaystyle \frac{\pi}{2}$. So the root test does work.
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    Senior Member Pinkk's Avatar
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    Isn't it safe to assume $\displaystyle sin(k) \ne 1$ because $\displaystyle k=1,2,3,...$, or in other words, $\displaystyle k$ is only ever an integer?

    Edit: redsoxfan beat me to it.
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    Quote Originally Posted by redsoxfan325 View Post
    $\displaystyle \sin k\neq 1~\forall k$ because $\displaystyle k\in\mathbb{N}$ and thus is never an odd multiple of $\displaystyle \frac{\pi}{2}$. So the root test does work.

    We can't conclude that, i thought the same thing but you have to prove it.


    It could come really close to 1. For example sin 55 = -0.999755

    As you go higher it gets closer to 1.

    But i would like to see some sound mathematical prove that sink =/= 1

    Ofcourse this is probably true but how close can it go.

    We would probably agree thaty 0.999999 = 1
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  9. #9
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by ╔(σ_σ)╝ View Post
    We can't conclude that, i thought the same thing but you have to prove it.


    It could come really close to 1. For example sin 55 = -0.999755

    As you go higher it gets closer to 1.

    But i would like to see some sound mathematical prove that sink =/= 1

    Ofcourse this is probably true but how close can it go.

    We would probably agree thaty 0.999999 = 1
    Assume $\displaystyle \exists~k$ such that $\displaystyle \frac{k}{\pi}\in\mathbb{Q}$ (i.e. $\displaystyle \frac{k}{\pi}=\frac{a}{b})$. Thus, by cross-multiplying and such, we have that $\displaystyle \pi = \frac{bk}{a}$, implying that $\displaystyle \pi\in\mathbb{Q}$, which is clearly false, creating a contradiction. Thus, $\displaystyle \nexists~k$ such that $\displaystyle \frac{k}{\pi}\in\mathbb{Q}$ and therefore $\displaystyle \sin k\neq 1~\forall k$.
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    Quote Originally Posted by redsoxfan325 View Post
    Assume $\displaystyle \exists~k$ such that $\displaystyle \frac{k}{\pi}\in\mathbb{Q}$ (i.e. $\displaystyle \frac{k}{\pi}=\frac{a}{b})$. Thus, by cross-multiplying and such, we have that $\displaystyle \pi = \frac{bk}{a}$, implying that $\displaystyle \pi\in\mathbb{Q}$, which is clearly false, creating a contradiction. Thus, $\displaystyle \nexists~k$ such that $\displaystyle \frac{k}{\pi}\in\mathbb{Q}$ and therefore $\displaystyle \sin k\neq 1~\forall k$.

    That does it.
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  11. #11
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    The root test is inconclusive here. Although it is true that $\displaystyle \frac{\sin k+2}3<1$ for all k, this fraction can get arbitrarily close to 1. Therefore $\displaystyle \limsup_{k\to\infty}\frac{\sin k+2}3=1$, and the root test does not cover this case.

    It certainly looks as though the series $\displaystyle \sum\Bigl(\frac{\sin k+2}3\Bigr)^k$ ought to converge, but I don't know of a test that would prove that. It's not even clear that the terms of this series converge to 0. We know that $\displaystyle \frac{\sin k+2}3$ can be very close to 1. But can it be so close to 1 that its k'th power is bounded away from 0, for some arbitrarily large k?
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  12. #12
    Super Member redsoxfan325's Avatar
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    I forgot that the root test was $\displaystyle \lim\sup a_n$; I had thought it was just $\displaystyle \lim a_n$. Graphing this seems to indicate that k gets large enough to counteract $\displaystyle \sin(k)$ approaching 1, but obviously that doesn't constitute a proof.
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  13. #13
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    That what i was getting at that before.

    But then someguy confused me saying that

    If $\displaystyle k = |\frac{n+1}{\frac{pi}{2}} | = 1 $

    But this is a contradition since K is an integer.


    But Opalg is right you can get really really close to 1, but how close?



    I think this question boils down to :
    Is there a finite number of time such that sink =~ 1 ?
    Last edited by Banned for attempted hacking; May 9th 2009 at 12:07 PM. Reason: gramma
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  14. #14
    Super Member redsoxfan325's Avatar
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    Well, I'm pretty sure that for any $\displaystyle \epsilon>0$, there will be infinitely many k such that $\displaystyle |1-\sin(k)|<\epsilon$, so I think the question really whether eventually k gets so large that $\displaystyle \sin^k(k)\to 0$.
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  15. #15
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    Quote Originally Posted by redsoxfan325 View Post
    Well, I'm pretty sure that for any $\displaystyle \epsilon>0$, there will be infinitely many k such that $\displaystyle |1-\sin(k)|<\epsilon$, so I think the question really whether eventually k gets so large that $\displaystyle \sin^k(k)\to 0$.
    sink oscillates so regardless of how small sink gets such that $\displaystyle sin^{k}k\to 0$. It eventually become 1 again.

    That why i say that is there must be a finite number of times that sink =1.
    If it happens infinitely many times then you are also saying that we have infinitely many +1's then it would follow that our series diverges.
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