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    I'm struggling to find what a(n) is in this question. The sequence from the values you get from part a) are:-

    a1=1
    a2=2/3
    a3=5/9
    a4=14/27

    I know that it must be something over 3^(n-1) but not sure what the numerator is, can anyone help?
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    Haven't read the question but that looks like \displaystyle \frac{1+3^{(1-n)}}{2} to me.
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    (Original post by As_Dust_Dances_)
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    I'm struggling to find what a(n) is in this question. The sequence from the values you get from part a) are:-

    a1=1
    a2=2/3
    a3=5/9
    a4=14/27

    I know that it must be something over 3^(n-1) but not sure what the nominator is, can anyone help?
    You don't need to find the nth term of the sequence explicitly to answer this question (and it'll be more taxing to do so just from the first few terms and guesswork but, if you're interested, google "solving linear recurrence relations" for an alternative method).

    Is there any specific part of the question that you cannot do? You haven't given many details as to where you're actually stuck. Assuming you've done the first part already, use the relation between a(n+1) and a(n) to do the induction in the second.
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    (Original post by As_Dust_Dances_)
    Nominator
    Numerator.
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    (Original post by As_Dust_Dances_)
    Name:  indt.png
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    I'm struggling to find what a(n) is in this question. The sequence from the values you get from part a) are:-

    a1=1
    a2=2/3
    a3=5/9
    a4=14/27

    I know that it must be something over 3^(n-1) but not sure what the nominator is, can anyone help?
    You don't need to find the nth term for this question. If you really want to, the sequence is:

    1 / 3^0
    (1 + 3^0) / 3^1
    (1 + 3^0 + 3^1) / 3^2
    (1 + 3^0 + 3^1 + 3^2) / 3^3

    You can use the sum of a geometric series formula to find the numerator part.
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    (Original post by Mr M)
    Numerator.
    Ouch, that's a bad mistake on my part to say I'm doing a Maths degree! D:
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    (Original post by Farhan.Hanif93)
    You don't need to find the nth term of the sequence explicitly to answer this question (and it'll be more taxing to do so just from the first few terms and guesswork but, if you're interested, google "solving linear recurrence relations" for an alternative method).

    Is there any specific part of the question that you cannot do? You haven't given many details as to where you're actually stuck. Assuming you've done the first part already, use the relation between a(n+1) and a(n) to do the induction in the second.
    for d) how would I show that a limit exists without a(n), or is it just case of putting large values into the sequence to show that it eventually converges to 0.5?
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    (Original post by As_Dust_Dances_)
    for d) how would I show that a limit exists without a(n), or is it just case of putting large values into the sequence to show that it eventually converges to 0.5?
    You have a decreasing sequence that is bounded below so…

    Then once you know the limit exists: it follows that you must have \displaystyle\lim_{n\to \infty} a_{n+1} = \displaystyle\lim_{n\to \infty} a_n =a where a is the limit you seek (observe that you wouldn't be able to do this had you not known that such an a exists in the first place). Can you see how to obtain an equation in terms of a to solve?
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    (Original post by Farhan.Hanif93)
    You have a decreasing sequence that is bounded below so…

    Then once you know the limit exists: it follows that you must have \displaystyle\lim_{n\to \infty} a_{n+1} = \displaystyle\lim_{n\to \infty} a_n =a where a is the limit you seek (observe that you wouldn't be able to do this had you not known that such an a exists in the first place). Can you see how to obtain an equation in terms of a to solve?
    so from this you would get 3a - 1 = a (is this correct?)
    giving a =1/2
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    (Original post by As_Dust_Dances_)
    so from this you would get 3a - 1 = a (is this correct?)
    giving a =1/2
    Indeed. But do you see why the limit exists? (should be in your notes somewhere but just in case…)
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    (Original post by Farhan.Hanif93)
    Indeed. But do you see why the limit exists? (should be in your notes somewhere but just in case…)
    I know why, but I'm not really sure how I would 'show' it exists. My Algebra lecturer explains things in much more difficult ways and so my notes look quite complicated to understand.. (sorry if I'm taking up your time!)
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    (Original post by As_Dust_Dances_)
    I know why, but I'm not really sure how I would 'show' it exists. My Algebra lecturer explains things in much more difficult ways and so my notes look quite complicated to understand.. (sorry if I'm taking up your time!)
    "Show that the limit exists" here is just a rephrasing of "show that the sequence converges". But you know that bounded, monotonic sequences converge (this is the bit I was referring to when I mentioned your notes) so, by showing that a_n is bounded below and decreasing, it follows that a_n converges and hence it is shown that the limit exists.
 
 
 
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