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Sequences

A sequence is defined by the iterative formula an+1=11−30an a_{n+1}= 11-\frac{30}{a_n} where n≥0 n\geq 0 and a0>0 a_0 > 0 .
Find the values of p p and q q such that if p<a0<q p<a_0 < q then the sequence is strictly increasing.
Find the limit L L of the sequence as n→∞ n \rightarrow \infty prove that an<L a_n < L for all n≥0 n\geq 0 .

This seems to be quite and awkward question and not too easy to go about it.
(edited 7 years ago)
Original post by Ano123
A sequence is defined by the iterative formula an+1=11−30an a_{n+1}= 11-\frac{30}{a_n} where n≥0 n\geq 0 and a0>0 a_0 > 0 .
Find the values of m m and n n such that if m<a0<n m<a_0 < n then the sequence is strictly increasing.
Find the limit L L of the sequence as n→∞ n \rightarrow \infty prove that an<L a_n < L for all n≥0 n\geq 0 .

This seems to be quite and awkward question and not too easy to go about it.


What have you tried?
Reply 2
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Ano123
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Original post by RDKGames
What have you tried?


Just replace all the a 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?
Original post by Ano123
Just replace all the a 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.
Original post by Ano123
Just replace all the a 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 a0a_0 is 0<a0<30110<a_0<\frac{30}{11} or a0>3011a_0>\frac{30}{11}. I'm not too great at these so I could be wrong.
Reply 5
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RichE
15
Original post by Ano123
A sequence is defined by the iterative formula an+1=11−30an a_{n+1}= 11-\frac{30}{a_n} where n≥0 n\geq 0 and a0>0 a_0 > 0 .
Find the values of p p and q q such that if p<a0<q p<a_0 < q then the sequence is strictly increasing.
Find the limit L L of the sequence as n→∞ n \rightarrow \infty prove that an<L a_n < L for all n≥0 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?
(edited 7 years ago)
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 a0a_0 is 0<a0<30110<a_0<\frac{30}{11} or a0>3011a_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

(edited 7 years ago)
Reply 7
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Ano123
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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<an<6 5<a_n < 6 the sequence is increasing. How would I prove that an<6 ∀n a_n < 6 \ \forall n ?
Original post by ValerieKR
The sequence is constant for both 5 and 6

Spoiler



Yeah you're right, I was thinking about it wrong.
Original post by RDKGames
Yeah you're right, I was thinking about it wrong.


The question isn't as clear as it could be
Original post by Ano123
How do you prove that the sequence converges? I have shown that if 5<an<6 5<a_n < 6 the sequence is increasing. How would I prove that an<6 ∀n a_n < 6 \ \forall n ?


You don't have to prove that an<6a_n < 6 necessarily. Your upper bound does not have to be your limit, it could be anything so long as ana_n is always less than or equal to it.
(edited 7 years ago)
Reply 11
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Zacken
22
Original post by Ano123
How do you prove that the sequence converges? I have shown that if 5<an<6 5<a_n < 6 the sequence is increasing. How would I prove that an<6 ∀n a_n < 6 \ \forall n ?


As above, with what 16Characters said, but also: induction.
Reply 12
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Ano123
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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 an<6 a_n <6
⇒1an>16 \Rightarrow \frac{1}{a_n} > \frac{1}{6}

⇒30an>5 \Rightarrow \frac{30}{a_n} > 5
⇒−30an<−5 \Rightarrow -\frac{30}{a_n} < -5
⇒11−30an<6 \Rightarrow 11-\frac{30}{a_n} < 6
⇒an+1<6 \Rightarrow a_{n+1} < 6 .
We know that a0<6 a_0 < 6 so an<6 ∀n a_{n} < 6 \ \forall n .

Is this really a sufficient proof?
(edited 7 years ago)
Original post by Ano123
I tried to use induction but the argument seemed quite weak.


What was your argument?
Reply 14
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Zacken
22
Original post by Ano123
I tried to use induction but the argument seemed quite weak.


Really? Pops out in a line for me: an<6⇒an+1=11−30an<11−306=6a_n < 6 \Rightarrow a_{n+1} = 11 - \frac{30}{a_n} < 11 - \frac{30}{6} = 6.
Reply 15
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Ano123
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Original post by ValerieKR
What was your argument?


Original post by Zacken
Really? Pops out in a line for me: an<6⇒an+1=11−30an<11−306=6a_n < 6 \Rightarrow a_{n+1} = 11 - \frac{30}{a_n} < 11 - \frac{30}{6} = 6.


Is what I put above ok?
Original post by Ano123
Is what I put above ok?


Yes
Reply 17
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Zacken
22
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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