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    A sequence is defined by the iterative formula  a_{n+1}= 11-\frac{30}{a_n} where  n\geq 0 and  a_0 > 0 .
    Find the values of  p and  q such that if  p<a_0 < q then the sequence is strictly increasing.
    Find the limit  L of the sequence as  n \rightarrow \infty prove that  a_n < L for all  n\geq 0 .

    This seems to be quite and awkward question and not too easy to go about it.
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    (Original post by Ano123)
    A sequence is defined by the iterative formula  a_{n+1}= 11-\frac{30}{a_n} where  n\geq 0 and  a_0 > 0 .
    Find the values of  m and  n such that if  m<a_0 < n then the sequence is strictly increasing.
    Find the limit  L of the sequence as  n \rightarrow \infty prove that  a_n < L for all  n\geq 0 .

    This seems to be quite and awkward question and not too easy to go about it.
    What have you tried?
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    (Original post by RDKGames)
    What have you tried?
    Just replace all the  a terms with L and solve for L. But this doesn't prove that L exists. The proof is the most difficult bit. Any pointers?
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    (Original post by Ano123)
    Just replace all the  a terms with L and solve for L. But this doesn't prove that L exists. The proof is the most difficult bit. Any pointers?
    An increasing sequence bounded above must converge.
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    (Original post by Ano123)
    Just replace all the  a terms with L and solve for L. But this doesn't prove that L exists. The proof is the most difficult bit. Any pointers?
    Not sure. I'm finding contradictions with that one. Since the limit is 6, it is usually approached from above it rather than below it regardless of whether a_0 is 0<a_0<\frac{30}{11} or a_0>\frac{30}{11}. I'm not too great at these so I could be wrong.
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    (Original post by Ano123)
    A sequence is defined by the iterative formula  a_{n+1}= 11-\frac{30}{a_n} where  n\geq 0 and  a_0 > 0 .
    Find the values of  p and  q such that if  p<a_0 < q then the sequence is strictly increasing.
    Find the limit  L of the sequence as  n \rightarrow \infty prove that  a_n < L for all  n\geq 0 .

    This seems to be quite and awkward question and not too easy to go about it.
    Drawing graphs of y=11-30/x and y=x may help - have you come across cobwebbing?
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    (Original post by RDKGames)
    Not sure. I'm finding contradictions with that one. Since the limit is 6, it is usually approached from above it rather than below it regardless of whether a_0 is 0<a_0<\frac{30}{11} or a_0>\frac{30}{11}. I'm not too great at these so I could be wrong.
    The sequence is constant for both 5 and 6
    Spoiler:
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    I haven't actually written anything down or worked it through properly but I suspect they are p and q, and you're supposed to take a0 as between them for that final limit part, in which case it would be strictly increasing but never exceed 6, else you could just make a0 10^10^10^10^G64 and the only sensible L would be 11, which is clearly not the case
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    (Original post by 16Characters....)
    An increasing sequence bounded above must converge.
    How do you prove that the sequence converges? I have shown that if  5<a_n < 6 the sequence is increasing. How would I prove that  a_n < 6 \ \forall n ?
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    (Original post by ValerieKR)
    The sequence is constant for both 5 and 6
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    I haven't actually written anything down or worked it through properly but I suspect they are p and q, and you're supposed to take a0 as between them for that final limit part, in which case it would be strictly increasing but never exceed 6
    Yeah you're right, I was thinking about it wrong.
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    (Original post by RDKGames)
    Yeah you're right, I was thinking about it wrong.
    The question isn't as clear as it could be
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    (Original post by Ano123)
    How do you prove that the sequence converges? I have shown that if  5<a_n < 6 the sequence is increasing. How would I prove that  a_n < 6 \ \forall n ?
    You don't have to prove that a_n < 6 necessarily. Your upper bound does not have to be your limit, it could be anything so long as a_n is always less than or equal to it.
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    (Original post by Ano123)
    How do you prove that the sequence converges? I have shown that if  5<a_n < 6 the sequence is increasing. How would I prove that  a_n < 6 \ \forall n ?
    As above, with what 16Characters said, but also: induction.
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    (Original post by Zacken)
    As above, with what 16Characters said, but also: induction.
    I tried to use induction but the argument seemed quite weak.
    I said that if  a_n <6
     \Rightarrow \frac{1}{a_n} > \frac{1}{6}

     \Rightarrow \frac{30}{a_n}  > 5
     \Rightarrow -\frac{30}{a_n} < -5
     \Rightarrow 11-\frac{30}{a_n} < 6
     \Rightarrow a_{n+1} < 6 .
    We know that  a_0 < 6 so  a_{n} < 6 \ \forall n .

    Is this really a sufficient proof?
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    (Original post by Ano123)
    I tried to use induction but the argument seemed quite weak.
    What was your argument?
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    (Original post by Ano123)
    I tried to use induction but the argument seemed quite weak.
    Really? Pops out in a line for me: a_n < 6 \Rightarrow a_{n+1} = 11 - \frac{30}{a_n} < 11 - \frac{30}{6} = 6.
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    (Original post by ValerieKR)
    What was your argument?
    (Original post by Zacken)
    Really? Pops out in a line for me: a_n < 6 \Rightarrow a_{n+1} = 11 - \frac{30}{a_n} < 11 - \frac{30}{6} = 6.
    Is what I put above ok?
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    (Original post by Ano123)
    Is what I put above ok?
    Yes
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    (Original post by Ano123)
    Is what I put above ok?
    Yes, that's what I did. Why d'you think it's not okay?
 
 
 
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