1. ## Cauchy Sequence

Not sure on the proof of this:

Prove directly from the definition that the sequence

$(\frac{sin(log(n^{2002})) + 5000}{2n+7})_n$

is cauchy

2. Originally Posted by Jason Bourne
Not sure on the proof of this:

Prove directly from the definition that the sequence

$(\frac{sin(log(n^{2002})) + 5000}{2n+7})_n$

is cauchy
I will state the following result. Your task is to prove it and show how it applies to your problem. Suppose that $\{a_n\},\{b_n\}$ are sequences such that $0\leq a_n \leq b_n$. Let $\{ b_n \}$ be Cauchy sequences and $\lim ~ a_n = \lim ~ b_n$. Prove that $\{ a_n \}$ is a Cauchy sequence.

3. Are you implying the General Principle of Convergence?

4. Originally Posted by Jason Bourne
Are you implying the General Principle of Convergence?
Frankly, I do not know what to make of the original question.
It is well know that a sequence of real numbers converges if and only if it is a Cauchy Sequence.
Take the above hint: Use basic comparison.
If you can show that your sequence converges then it is Cauchy.

5. Originally Posted by Plato
Frankly, I do not know what to make of the original question.
It is well know that a sequence of real numbers converges if and only if it is a Cauchy Sequence.
Take the above hint: Use basic comparison.
If you can show that your sequence converges then it is Cauchy.
I also do not know why his professor would not let him the underlined theorem. Unless, the professor does not want to use completeness. Since completeness does not work in arbitrary metric spaces.

6. I don't know what time it is where you guys live but I'm going to sleep. Thanks anyway.

7. Originally Posted by ThePerfectHacker
Unless, the professor does not want to use completeness. Since completeness does not work in arbitrary metric spaces.
I agree. But in arbitrary metric if a sequence converges then it is a Cauchy Sequence. (The other way requires completeness.) Thus if he shows by comparison the sequence converges it is Cauchy.

8. Originally Posted by Plato
I agree. But in arbitrary metric if a sequence converges then it is a Cauchy Sequence. (The other way requires completeness.) Thus if he shows by comparison the sequence converges it is Cauchy.
Yes ladies and gentlemen (I make no unjustified assumptions here). All of this is what I thought too, but didn't quite feel confident enough with my analysis to suggest it. I was interested to see what others thought before jumping in. A combination of comparison and then the definition would seem to satisfy the boundary conditions.

I quite agree with the puzzle of why his prof won't let the theorem that

a sequence of real numbers converges if and only if it is a Cauchy Sequence

be used. Using this theorem was my first thought.

Unless his prof wants them to practice the whole epsilon thing .....

I suspect the whole metric thing raised by TPH is probably a red herring in this case ......

9. Originally Posted by mr fantastic
Yes ladies and gentlemen (I make no unjustified assumptions here). All of this is what I thought too, but didn't quite feel confident enough with my analysis to suggest it. I was interested to see what others thought before jumping in. A combination of comparison and then the definition would seem to satisfy the boundary conditions.

I quite agree with the puzzle of why his prof won't let the theorem that

a sequence of real numbers converges if and only if it is a Cauchy Sequence

be used. Using this theorem was my first thought.

Unless his prof wants them to practice the whole epsilon thing .....

I suspect the whole metric thing raised by TPH is probably a red herring in this case ......
Indeed, I think the question wanted proof from the definition of a Cauchy sequence but it does not seem neccessary really.