# Does my proof work?

1. Sep 21, 2008

### Unassuming

Does my "proof" work?

n\in \mathbb{N}
\).

Last edited: Sep 21, 2008
2. Sep 21, 2008

### xaos

Re: Does my "proof" work?

not diverging may not be equivalent to converging. it may be merely bounded but oscillating. so i would contradict boundedness rather convergence. maybe by showing that the difference between the nth and the n+k is larger than any bound.

3. Sep 21, 2008

### morphism

Re: Does my "proof" work?

But such a sequence is certainly diverging / not converging.

About the proof presented in the OP: isn't your N_2 redundant? After all, n+1 > n >= N_1. Other than this minor quibble, your proof is fine.

4. Sep 21, 2008

### Unassuming

Re: Does my "proof" work?

I thought something was wrong with my proof. I was wondering whether N_2 was redundant. I can repeat the definition using N_1 again, or N.

I was wondering about the contradiction though. Does it really contradict? I proved the opposite based off the assumption, therefore I can assume the original is correct?

5. Sep 21, 2008

### morphism

Re: Does my "proof" work?

You showed that for any convergent sequence (a_n), the quantity |a_{n+1} - a_n| can be made arbitrarily small (by choosing sufficiently large n). This means that a sequence for which this cannot happen is divergent.

6. Sep 21, 2008

### Unassuming

Re: Does my "proof" work?

ahh, thank you

7. Sep 21, 2008

### xaos

Re: Does my "proof" work?

yeah my first guess is usually wrong. you would think that divergence is always unbounded.