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Thread: Finding limit of S

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    Finding limit of S

    Let


    then find
    Mahan Ghobadi

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    Re: Finding limit of S

    I got 2/3. take out the 1/9 and you get Sn as 1/9(1/4 x2/5x3/6...)
    the numerators and denominators of this cancel except for the first 1x2x3 and hence
    sn=1/9x6=2/3
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    Re: Finding limit of S

    Quote Originally Posted by Kingom View Post
    I got 2/3. take out the 1/9 and you get Sn as 1/9(1/4 x2/5x3/6...)
    the numerators and denominators of this cancel except for the first 1x2x3 and hence
    sn=1/9x6=2/3
    Unfortunately, the answer is not .
    You have an interesting idea but when it gets to infinite product or sum, weird things happen. I believe, you can't in general use telescoping method for infinite sums or products.
    Consider this example:

    let
    (By Leibniz test s is finite, )

    but dividing s by 2 we get the series, t:




    by REARRANGING the terms we get :



    which is a contradiction. it turns out rearranging the terms in an infinite series requires more care than when we deal with finite series.

    So I suggest, you get the general form of then use limit to find the answer.
    Hint: use the same cancellation pattern but for finite n.
    Last edited by Mahan1; 20 Feb 2017 at 12:43 AM.
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    Re: Finding limit of S

    Quote Originally Posted by Mahan1 View Post
    Unfortunately, the answer is not .
    You have an interesting idea but when it gets to infinite product or sum, weird things happen. I believe, you can't in general use telescoping method for infinite sums or products.
    Consider this example:

    let
    (By Leibniz test s is finite, )

    but dividing s by 2 we get the series, t:




    by REARRANGING the terms we get :



    which is a contradiction. it turns out rearranging the terms in an infinite series requires more care than when we deal with finite series.

    So I suggest, you get the general form of then use limit to find the answer.
    Hint: use the same cancellation pattern but for finite n.
    It will depend on whether the infinite sum converges or not.
    If it diverges, then possibly you can rearrange it to make it converge to whatever value you want, including contradictions in some cases.

    In this case, a quick test for a infinite series whether it converges, is does the nth term tend towards 0.
    (clearly if it is an alternating series, there is a different test)
    Last edited by dan964; 20 Feb 2017 at 8:07 AM.
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    Re: Finding limit of S

    Here's my thinking on it. I got the same answer as Kingom but I haven't done many infinite product questions...

    Last edited by BenHowe; 20 Feb 2017 at 4:05 PM.
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    Re: Finding limit of S

    Quote Originally Posted by BenHowe View Post
    Here's my thinking on it. I got the same answer as Kingom but I haven't done many infinite product questions...

    You are making the same mistake.
    instead of infinity you should find for finite k that means


    then
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    Re: Finding limit of S

    An interesting extension is:

    Last edited by seanieg89; 20 Feb 2017 at 1:40 PM.
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    Re: Finding limit of S

    Quote Originally Posted by Mahan1 View Post
    You are making the same mistake.
    instead of infinity you should find for finite k that means


    then
    So k is just any random number that satisfies ? And if so you need to do this because it doesn't make sense for k to be infinity...? Don't quite understand what you mean
    Last edited by BenHowe; 20 Feb 2017 at 3:54 PM.
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    Re: Finding limit of S

    Quote Originally Posted by BenHowe View Post
    So k is just any random number that satisfies ?
    Last edited by InteGrand; 20 Feb 2017 at 3:58 PM.

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    Re: Finding limit of S

    Quote Originally Posted by seanieg89 View Post
    An interesting extension is:

    Is it always ? I don't know how to explain it for all cases but I just tried because I think it satisfies the above condition and I got , so I thought it'd be ?
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    Re: Finding limit of S

    Quote Originally Posted by InteGrand View Post
    But I don't quite understand the difference between and .
    Last edited by BenHowe; 20 Feb 2017 at 4:52 PM.
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    Re: Finding limit of S

    Quote Originally Posted by BenHowe View Post
    But I don't quite understand the difference between and . Sorry if this seems stupid my teacher hadn't taught ext1/2 maths before so I did my best, but I have some gaps.
    They are the same, but when you found the infinite product with your method, you changed around the order of multiplication, and for infinite products, this can change the answer (like how it can for infinite sums (see: https://en.wikipedia.org/wiki/Riemann_series_theorem); note that an infinite product of positive terms will converge to a non-zero number if and only if the infinite sum of logs of terms converges to a real number (and "diverges" to 0 if the infinite sum of logs of terms diverges to negative infinity). You can find more information at the Wikipedia page for infinite products: https://en.wikipedia.org/wiki/Infinite_product .).

    And don't worry, infinite products aren't really in the HSC syllabus.
    Last edited by InteGrand; 20 Feb 2017 at 5:12 PM.

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    Re: Finding limit of S

    Quote Originally Posted by BenHowe View Post
    Is it always ? I don't know how to explain it for all cases but I just tried because I think it satisfies the above condition and I got , so I thought it'd be ?
    It is always zero when the degree of the denominator exceeds the degree of the numerator by exactly one. (As in the original question in this thread, although to prove this more general statement you do not have telescoping at your disposal.)

    Your second product does not satisfy P(n) < Q(n) for all n, which is a hypothesis of the question. (If one of the factors is zero, then certainly the infinite product tends to zero.)
    Last edited by seanieg89; 20 Feb 2017 at 10:34 PM.
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    Re: Finding limit of S

    Quote Originally Posted by seanieg89 View Post
    It is always zero when the degree of the denominator exceeds the degree of the numerator by exactly one. (As in the original question in this thread, although to prove this more general statement you do not have telescoping at your disposal.)

    Your second product does not satisfy P(n) < Q(n) for all n, which is a hypothesis of the question. (If one of the factors is zero, then certainly the infinite product tends to zero.)
    Sorry to bother you, would you be able to post a proof then?
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    Re: Finding limit of S

    Quote Originally Posted by InteGrand View Post
    They are the same, but when you found the infinite product with your method, you changed around the order of multiplication, and for infinite products, this can change the answer (like how it can for infinite sums (see: https://en.wikipedia.org/wiki/Riemann_series_theorem); note that an infinite product of positive terms will converge to a non-zero number if and only if the infinite sum of logs of terms converges to a real number (and "diverges" to 0 if the infinite sum of logs of terms diverges to negative infinity). You can find more information at the Wikipedia page for infinite products: https://en.wikipedia.org/wiki/Infinite_product .).

    And don't worry, infinite products aren't really in the HSC syllabus.
    So what you're telling me is that ? Is there a rule kinda like a product of matrices going on here?
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    Re: Finding limit of S

    By the degree condition nP(n)/Q(n) tends to some positive constant limit (the ratio of leading coefficients.) Let's call this 2c.

    Then for sufficiently large n (say n > K), we have nP(n)/Q(n) > c, and (1-c/n) > 0.

    Then for N > k:

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    Re: Finding limit of S

    Quote Originally Posted by seanieg89 View Post
    By the degree condition nP(n)/Q(n) tends to some positive constant limit (the ratio of leading coefficients.) Let's call this 2c.

    Then for sufficiently large n (say n > K), we have nP(n)/Q(n) > c, and (1-c/n) > 0.

    Then for N > k:

    You lost me... Why did you go to logs and sigma?
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    Re: Finding limit of S

    Quote Originally Posted by BenHowe View Post
    You lost me... Why did you go to logs and sigma?
    It's a common trick to convert infinite products into infinite sums which are in some ways nicer to work with.

    There were also two facts used that I didn't bother writing proofs of (you should try to make sure you understand why they are true):

    1. log(1+x) =< x for all x > -1 where these things are defined.

    This is an MX2 level application of calculus.

    2. The harmonic series (1+1/2+1/3+...) diverges.

    This can be done in many many different ways. If you like calculus, then you can bound the partial sums of this series below by an integral involving 1/x. The resulting quantity will grow like log(x) and hence diverge to +infinity.
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    Re: Finding limit of S

    Quote Originally Posted by BenHowe View Post
    So what you're telling me is that ? Is there a rule kinda like a product of matrices going on here?



























    Last edited by Mahan1; 21 Feb 2017 at 1:04 PM.
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    Re: Finding limit of S

    Quote Originally Posted by Mahan1 View Post
    You are making the same mistake.
    instead of infinity you should find for finite k that means


    then
    The problem is the for infinite products doesn't hold.
    same goes for infinite sums.

    A similar case is swapping sums and integrals, which requires the following theorem called the Dominated Convergence Theorem...
    https://en.wikipedia.org/wiki/Domina...rgence_theorem
    (sorry was lazy and linked Wikipedia)
    and refer to the answer to this question:
    http://math.stackexchange.com/questi...e-interchanged

    In terms of what Mahan is alluding to in his above post, is defining an infinite series or product, in terms of its "partial sums" or partial products. We can consider these partial sums/partial products in turn, as a infinite sequence.

    Now infinite sequences are fairly easy to grapple with once we understand function and limits and that a sequence
    is a function.
    Note the codomain (place where we want the function to end up in), of which the range is a part of can be or (although it does become a bit interesting when considering limits in the later), or some portion of either e.g. the rational numbers.


    And hence then we can take a limit on the function, that is the sequence of partial sums/products as ....
    Last edited by dan964; 26 Feb 2017 at 10:08 PM.
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    Re: Finding limit of S

    Quote Originally Posted by seanieg89 View Post
    It's a common trick to convert infinite products into infinite sums which are in some ways nicer to work with.

    There were also two facts used that I didn't bother writing proofs of (you should try to make sure you understand why they are true):

    1. log(1+x) =< x for all x > -1 where these things are defined.

    This is an MX2 level application of calculus.

    2. The harmonic series (1+1/2+1/3+...) diverges.

    This can be done in many many different ways. If you like calculus, then you can bound the partial sums of this series below by an integral involving 1/x. The resulting quantity will grow like log(x) and hence diverge to +infinity.
    when using the squeeze theorem what was the other function/limit you used??
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    Re: Finding limit of S

    Quote Originally Posted by dan964 View Post
    when using the squeeze theorem what was the other function/limit you used??
    P_N is always positive by definition, so we just need to bound it above by something that tends to zero.

    In the notation I introduced earlier, this upper bound is:

    C/exp(h_k-h_N)

    where h_n = 1 + 1/2 + ... + 1/n denotes the n-th partial sum of the harmonic series, and C and k are constants.

    (k and K mean the same thing, I changed my mind about capitalisation and couldn't be bothered to fix all of them.)
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