I answered a question but I don't know if my answer is correct so I would appreciate if someone can confirm it. Thank you.

By the way I did some mistake when I wrote it. its a_{n+1}-a_{n}

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- Dec 13th 2013, 12:03 PMdavidciprutNeed confirmation.
I answered a question but I don't know if my answer is correct so I would appreciate if someone can confirm it. Thank you.

By the way I did some mistake when I wrote it. its a_{n+1}-a_{n} - Dec 13th 2013, 12:21 PMromsekRe: Need confirmation.
suppose

$\displaystyle a_n=(-1)^n$

then

$\displaystyle \forall n (a_{n+1}+a_n)=0$

certainly $\displaystyle a_n$ doesn't converge. It is however bounded. I don't think you can play the same trick if {a} is unbounded. I'd have to think about it. - Dec 13th 2013, 12:38 PMdavidciprutRe: Need confirmation.
Hmmm, I didn't think about that, I thought if it didn't work I had to come to a contradiction algebrically from the definition of CAUCHY or something .. So in this case I have to give a counter-example ?

- Dec 13th 2013, 12:48 PMromsekRe: Need confirmation.
Well just answer the questions.

Does this mean $\displaystyle a_n$ converges to any point? No, we showed that.

Does this mean $\displaystyle a_n$ is bounded? We haven't shown it but I suspect so. Show this. - Dec 13th 2013, 12:59 PMromsekRe: Need confirmation.
Looks like I'm wrong.

$\displaystyle a_n=(-1)^n log(n)$

clearly $\displaystyle a_n$ is unbounded.

$\displaystyle a_n+a_{n+1}=(-1)^n log(n) + (-1)^{n+1} log(n+1)$

let n be even

$\displaystyle a_n+a_{n+1}=log(n)-log(n+1)=log(\frac{n}{n+1})\rightarrow log(1)=0$

similar if n is odd

So you can manage to make the sum of adjacent elements converge to 0 even while the sequence itself is unbounded. - Dec 13th 2013, 01:12 PMSlipEternalRe: Need confirmation.
Let $\displaystyle b_n = a_{n+1} + a_n$. According to the question, $\displaystyle \lim_{n \to \infty} b_n = 0$. As romsek showed, the convergence of $\displaystyle b_n$ has nothing to do with the convergence of $\displaystyle a_n$. But, we can use $\displaystyle b_n$ to look at properties of $\displaystyle a_n$.

Check out $\displaystyle \lim_{n \to \infty} (b_{n+1} - b_n)$. By limit laws, the limit of the difference is the difference of the limits, so long as both limits exist. Since both do, we know that $\displaystyle \lim_{n \to \infty} (b_{n+1} - b_n) = \lim_{n \to \infty} (a_{n+2} - a_n) = 0$. You can do the same for $\displaystyle \lim_{n \to \infty} (b_{n+2} - b_{n+1}) = \lim_{n \to \infty} (a_{n+3} - a_{n+1}) = 0$. So, if you define sequences $\displaystyle c_n = a_{2n}, d_n = a_{2n+1}$, perhaps you can show that both of these sequences are Cauchy (using the previous results), and hence they both converge. Since convergent sequences are bounded, the original sequence must be bounded, as well. - Dec 13th 2013, 01:12 PMdavidciprutRe: Need confirmation.
Hmmmm, maybe we can take the sequence a

_{n}=n , so we can say that a_{n+1-}a_{n}converges to 1? but a_{n}isn't bounded? Is this possible? i mean i know infinite minus infinite in limits is a problem but the distance between them should be 1 and it shouldn't be a problem right?(I got confused , this is completely wrong - Dec 13th 2013, 01:19 PMSlipEternalRe: Need confirmation.
- Dec 13th 2013, 01:20 PMdavidciprutRe: Need confirmation.
Why are you looking at the sum? it says the difference between them converges to 0

- Dec 13th 2013, 01:27 PMdavidciprutRe: Need confirmation.
So when I posted I wrote the sum between them but I corrected it by writing a note on the post, did you guys read it ?

- Dec 13th 2013, 01:31 PMdavidciprutRe: Need confirmation.
So this is the correct one , although I corrected myself when I submitted the thread, sorry for confusing

- Dec 13th 2013, 02:47 PMromsekRe: Need confirmation.
say what? This is an entirely different problem.

In this case it's just a Cauchy sequence that clearly converges to a point and is thus bounded.

This proof must be on the web in a dozen places or so. Here.