haXadecimal
Aug6-04, 11:40 AM
Hi, I have to prove that an oscillating sequence converges, I am having some difficulty with the proof.
The sequence is c_{n+1} = \frac{1}{1+c_{n}} , c_{1} = 1
So, I've calculated the first few terms and have seen that the sequence oscillates. I know that I need to prove:
1) The differences alternate in sign.
2) The absolute differences decrease.
3) The absolute differences approach 0.
I have proved 1, using:
c_{n+1} - c_{n} = \left(\frac{1}{1+c_{n}}\right) - \left(\frac{1}{1+c_{n-1}}\right)
=\frac{1+c_{n-1}-1-c_{n}}{1+c_{n-1}+c_{n}+c_{n-1}c_{n}}
=\frac{-(c_{n}-c_{n-1})}{1+c_{n-1}+c_{n}=c_{n-1}c_{n}}
And since all terms are positive, the denomenator will be positive and the difference between two terms with alternate in sign from the difference between the previous two terms.
I now am having trouble proving 2 and 3. I'm not exactly sure what to do; the example in my book is not very helpful.So far I have:
|c_{n+1}-c_{n}| < |c_{n} - c_{n-1}|
but that's not much... If anyone could help, that would be great!! Thanks!
The sequence is c_{n+1} = \frac{1}{1+c_{n}} , c_{1} = 1
So, I've calculated the first few terms and have seen that the sequence oscillates. I know that I need to prove:
1) The differences alternate in sign.
2) The absolute differences decrease.
3) The absolute differences approach 0.
I have proved 1, using:
c_{n+1} - c_{n} = \left(\frac{1}{1+c_{n}}\right) - \left(\frac{1}{1+c_{n-1}}\right)
=\frac{1+c_{n-1}-1-c_{n}}{1+c_{n-1}+c_{n}+c_{n-1}c_{n}}
=\frac{-(c_{n}-c_{n-1})}{1+c_{n-1}+c_{n}=c_{n-1}c_{n}}
And since all terms are positive, the denomenator will be positive and the difference between two terms with alternate in sign from the difference between the previous two terms.
I now am having trouble proving 2 and 3. I'm not exactly sure what to do; the example in my book is not very helpful.So far I have:
|c_{n+1}-c_{n}| < |c_{n} - c_{n-1}|
but that's not much... If anyone could help, that would be great!! Thanks!