Converrgence of oscillating sequence

[haXadecimal]

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!

[arildno]

Note that:
(1+c_{n})(1+c_{n-1})=2+c_{n-1}
By substituting c_{n}=\frac{1}{1+c_{n-1}}

[haXadecimal]

Thank you! Ok, now I have:


|c_{n+1}-c_{n}| = \frac{|c_{n}-c_{n-1}|}{2+c_{n-1}}


And since all terms are positive, 2+c_{n-1} will be positive, and each absolute difference will be a fraction of the previous absolute difference. Therefore they are decreasing and they will approach 0 as n apporaches infinity. Is that enough to prove this by just saying this? Thanks!

(Is there any way to change the title of the thread? I made a typo :redface: )

[arildno]

Technically, I guess you should prove that the gained relations imply that we've got a Cauchy sequence, and hence, that the sequence converges (depends on what you may take as granted)